358. Some properties of the remarkable lines in a triangle.

by Virgil Nicula, Oct 15, 2012, 10:47 AM

P6 (Ruben Dario). Let $ABCD$ be a rectangle. Let $w_1=\mathbb C\left(O_1,R\right)$ be a circle what is tangent to $AD$ at $A$ and passes

through $C.$ Let $w_2=\mathbb C\left(O_2,r\right)$ be a circle what is tangent to $CD$ at $C$ and passes through $A.$ Show that $\frac 1{R^2}+\frac 1{r^2}=\frac 4{AC^2}.$

Proof. Denote $AB=b$ and $AD=a.$ The circle $w_1$ has the diameter $[AE],$ where $E\in AB$ so that $CE\perp CA.$ Denote $BE=x,$ i.e. $\boxed{bx=a^2\ \wedge\ b+x=2R}\ (1).$

The circle $w_2$ has the diameter $[CF],$ where $F\in CB$ so that $AF\perp AC.$ Denote $BF=y,$ i.e. $\boxed{ay=b^2\ \wedge\ a+y=2r}\ (2).$ Denote $O\in AC\cap BD$ and

$G\in O_1O_2\cap AD.$ Prove easily that $O\in O_1O_2$ and $AG=AO_2=r.$ Observe that $AO$ is the $A$ -altitude in the $A$ -right triangle $O_1AG.$ Apply an well known property $:$

$"(\forall )\ A$ -right $\triangle ABC$ there is the identity $\frac 1{b^2}+\frac 1{c^2}=\frac 1{h^2},$ where $h$ is the length of the $A$ -altitude $".$ Hence $\frac 1{AO_1^2}+\frac 1{AG^2}=\frac 1{AO^2},$ i.e. $\frac 1{R^2}+\frac 1{r^2}=\frac 4{AC^2}.$

Remark. From the relations $(1),$ $(2)$ and $a^2+b^2=AC^2$ obtain directly the required relation $\frac 1{R^2}+\frac 1{r^2}=\frac 4{AC^2}.$

P7 (M. O. Sanchez). Let an $A$ -right $\triangle ABC$ and $H\in (BC)$ so that $AH\perp BC.$ Denote $:$ the incenter $I_1$ and the $A$ -excenter $E_1$ of

$\triangle ABH\ ;$ the incenter $I_2$ and the $A$ -excenter $E_2$ of $\triangle ACH.$ Prove that the area $\left[I_1I_2E_2E_1\right]=S,$ where $S$ is the area of $\triangle ABC.$

Proof. $\frac {HI_1}{r_1}=\frac {HE_2}{\rho_2}=$ $\frac {HI_2}{r_2}=\frac {HE_1}{\rho_1}=$ $\sqrt 2$ and $\left\{\begin{array}{ccccc} \triangle ABH\sim \triangle CBA & \implies & \frac {r_1}r=\frac ca=\frac {\rho_1}{r_c}\\\\ \triangle ACH\sim \triangle BCA & \implies & \frac {r_2}r=\frac ba=\frac {\rho_2}{r_b}\end{array}\right\|\ (*)\$ $\implies$ $\left[I_1I_2E_2E_1\right]=\frac 12\cdot I_1E_2\cdot I_2E_1=$

$\frac 12\cdot \left(HI_1+HE_2\right)\left(HI_2+HE_1\right)=$ $\left(r_1+\rho_2\right)\left(r_2+\rho _1\right)=$ $\left(\frac ca\cdot r+\frac ba\cdot r_b\right)\left(\frac ba\cdot r+\frac ca\cdot r_c\right)=$ $\frac 1{a^2}\cdot \left(cr+br_b\right)\left(br+cr_c\right)=$

$\frac 1{a^2}\cdot \left(bcr^2+c^2rr_c+b^2rr_b+bcr_br_c\right)=$ $\frac 1{a^2}\cdot\left[bc(s-a)^2+c^2(s-a)(s-b)+b^2(s-a)(s-c)+bcs(s-a)\right]=$

$\frac {s-a}{a^2}\cdot\left[bc(s-a)+c^2(s-b)+b^2(s-c)+bcs\right]=$ $\frac {s-a}{a^2}\cdot\left[bc(b+c)+s\left(b^2+c^2\right)-bc(b+c)\right]=s(s-a)=r_ar=S=\left[I_1I_2E_2E_1\right].$

P8 (Ruben Dario). Let $\triangle ABC$ with the incircle $w=\mathbb C(I,r)$ which touches $BC,$ $CA$ and $AB$ at $D,$ $E$ and $F.$ Denote $:\ P\in (AF),$ $S\in (AE)$ so that $PS\parallel BC$

and is tangent to $w;$ $N\in BI\cap PS,$ $R\in BC\cap AN$ and $M\in BC$ so that $MN$ is tangent to $w,$ i.e. $BPNM$ is a rhombus. Prove that $\frac {CM}{CR}=\tan^2\frac B2.$

$\begin{array}{c} \underline{\mathrm{PROOF}}\\\\ \mathrm{(metric)}\end{array}\ :\ \left\|\begin{array}{cccc} \frac {PS}{BC}=\frac {h_a-2r}{h_a}=\frac {ah_a-2ar}{ah_a}=\frac {2sr-2ar}{2sr}=\frac {s-a}s & \implies & \boxed{PS=\frac {a(s-a)}s} & (1)\\\\ BM=\frac {IB^2}{BD}=\frac {\frac {ac(s-b)}s}{s-b}=\frac {ac}s & \implies & \boxed{BM=\frac {ac}s} & (2)\\\\ DM=\frac {ID^2}{BD}=\frac {r^2}{s-b}=\frac {(s-a)(s-c)}s & \implies & \boxed{DM=\frac {(s-a)(s-c)}s} & (3)\\\\ CM=CD-DM\ \stackrel{(3)}{=}\ (s-c)-\frac {(s-a)(s-c)}s=\frac {a(s-c)}s & \implies & \boxed{CM=\frac {a(s-c)}s} & (4)\\\\ SN=PN-PS=BM-PS\ \stackrel{(1\wedge 2)}{=}\ \frac {ac}s-\frac {a(s-a)}s=\frac {a(a+c-s)}s=\frac {a(s-b)}s & \implies & \boxed{SN=\frac {a(s-b)}s} & (5)\\\\ CR=SN\cdot \frac {h_a}{h_a-2r}\ \stackrel{(1\wedge 5)}{=}\ \ldots =\frac {a(s-b)}s\cdot \frac s{s-a}=\frac {a(s-b)}{s-a} & \implies & \boxed{CR=\frac {a(s-b)}{s-a}} & (6)\\\\ \frac {CM}{CR}=\frac {a(s-c)}s\cdot\frac {s-a}{a(s-b)}=\frac {(s-a)(s-c)}{s(s-b)}=\tan^2\frac B2 & \implies & \boxed{\boxed{\frac {CM}{CR}\ \stackrel{4\wedge 6}{=}\ \tan^2\frac B2}} & (7)\end{array}\right\|\ \mathrm{Remark\ :}\ \boxed{\frac {PS}{s-a}=\frac {SN}{s-b}=\frac {CM}{s-c}=\frac as}\ (*)\ .$

P9 (CeuAzul). Let $\triangle ABC$ with centroid $G$ and the circumcircle $w.$ Let $\left\{\begin{array}{ccc} D\in AA\cap BC & ; & \{A,L\}=\{A,G\}\cap w\\\\ E\in DG\cap AB & ; & F\in DG\cap AC\end{array}\right\| .$ Prove that $\widehat{ALE}\equiv\widehat{ALF}.$ Very nice !

Proof (metric). Denote $:\ \frac {EB}{EA}=m,$ $\frac {FC}{FA}=n\ ;$ midpoint $M$ of $[BC]\ ;\ X\in LE\cap BC,$ $Y\in LF\cap BC\ ;$ diameter $[AS]$ of $w\ ;\ Z\in LS\cap BC.$ Apply the

Cristea's relation $:$ $\frac {EB}{EA}\cdot MC+\frac {FC}{FA}\cdot MB=\frac {GM}{GA}\cdot BC\iff$ $\boxed{m+n=1}\ (*)$ and $\frac {LA}{LM}=1+\frac{MA}{ML}=$ $1+\frac{MA^2}{MA\cdot ML}=$ $1+\frac{MA^2}{MB\cdot MC}=$

$1+\frac{4m_a^2}{a^2}$ $\implies$ $\boxed{\frac {LA}{LM}=\frac{2\left(b^2+c^2\right)}{a^2}}\ (1).$ Suppose w.l.o.g. that $b>c$ and denote $K\in BC$ so that $AK\perp BC.$ Apply the Menelaus' theorem to the transversals $:$

$\blacktriangleright\ \overline{DEF}/ABC\ ;\ \frac {DB}{DC}\cdot \frac {FC}{FA}\cdot \frac {EA}{EB}=1$ $\iff$ $\frac {c^2}{b^2}\cdot\frac nm=1$ $\stackrel{(*)}{\iff}$ $\frac m{c^2}=\frac n{b^2}=\frac 1{b^2+c^2}$ $\odot\begin{array}{cc} \nearrow & m=\frac {EB}{EA}=\frac {c^2}{b^2+c^2}\implies\frac {EB}{c^2}=\frac {EA}{b^2+c^2}=\frac {c}{b^2+2c^2}\\\\ \searrow & n=\frac {FC}{FA}=\frac {b^2}{b^2+c^2}\implies\frac {FC}{b^2}=\frac {FA}{b^2+c^2}=\frac {b}{2b^2+c^2}\end{array}\ (2)\ .$

$\blacktriangleright\ \overline{LXE}/ABM\ :\ \frac {LM}{LA}\cdot\frac {EA}{EB}\cdot\frac {XB}{XM}=1$ $\implies$ $\frac {XB}{XM}=$ $\frac {EB}{EA}\cdot \frac {LA}{LM}\ \stackrel{(1\wedge 2)}{=}\ \frac {2c^2}{a^2}$ $\implies$ $\frac {XB}{XM}=\frac {2c^2}{a^2}$ $\implies$ $\frac {XB}{2c^2}=\frac {XM}{a^2}=\frac {\frac a2}{a^2+2c^2}$ $\implies$

$XM=\frac {a^3}{2\left(a^2+2c^2\right)}\ (3)\ .$

$\blacktriangleright\ \overline{LYF}/ACM\ :\ \frac {LM}{LA}\cdot\frac {FA}{FC}\cdot\frac {YC}{YM}=1$ $\implies$ $\frac {YC}{YM}=$ $\frac {FC}{FA}\cdot \frac {LA}{LM}\ \stackrel{(1\wedge 2)}{=}\ \frac {2b^2}{a^2}$ $\implies$ $\frac {YC}{YM}=\frac {2b^2}{a^2}$ $\implies$ $\frac {YC}{2b^2}=\frac {YM}{a^2}=\frac {\frac a2}{a^2+2b^2}$ $\implies$

$YM=\frac {a^3}{2\left(a^2+2b^2\right)}\ (4)\ .$ From $(3)$ and $(4)$ obtain that $\boxed{\frac {MX}{MY}=\frac {a^2+2b^2}{a^2+2c^2}}\ (5)\ .$ Observe that $AKLZ$ is inscribed in the circle with the diameter $[AZ].$

Hence $MZ\cdot MK=MA\cdot ML=$ $MB\cdot MC$ $\iff$ $MZ\cdot \frac {\left(b^2-c^2\right)}{2a}=$ $\left(\frac a2\right)^2$ $\iff$ $\boxed{MZ=\frac {a^3}{2\left(b^2-c^2\right)}}\ (6)\ .$

Thus, $\left\{\begin{array}{ccccc} ZX=MZ+MX=\frac {a^3}{2\left(b^2-c^2\right)}+\frac {a^3}{2\left(a^2+2c^2\right)} & \implies & ZX=\frac {a^3\left(a^2+b^2+c^2\right)}{2\left(a^2+2c^2\right)\left(b^2-c^2\right)}\\\\ ZY=MZ-MY=\frac {a^3}{2\left(b^2-c^2\right)}-\frac {a^3}{2\left(a^2+2b^2\right)} & \implies & ZY=\frac {a^3\left(a^2+b^2+c^2\right)}{2\left(a^2+2b^2\right)\left(b^2-c^2\right)}\end{array}\right\|$ $\implies$ $\boxed{\frac {ZX}{ZY}=\frac {a^2+2b^2}{a^2+2c^2}}\ (7)\ .$

Therefore, $(6)\wedge (7)\implies\ \frac {MX}{MY}=\frac {ZX}{ZY}\iff$ the division $\{X,Y;M,Z\}$ is harmonic. Hence $LZ\perp LM$ $\iff$ $\widehat {MLX}\equiv\widehat{MLY}\iff$ $\widehat{ALE}\equiv\widehat{ALF}.$

Remark. Apply the well known relation $\frac {GE}{GF}=$ $\frac {MB}{MC}\cdot \frac {AE}{AB}\cdot \frac {AC}{AF}=$ $\frac {b^2+c^2}{b^2+2c^2}\cdot\frac {2b^2+c^2}{b^2+c^2}$ $\implies$ $\boxed{\frac {GE}{GF}=\frac {2b^2+c^2}{b^2+2c^2}}\ .$

Analogously (verify !) $\frac {MX}{MY}=\frac {GE}{GF}\cdot\frac {LX}{LE}\cdot\frac {LF}{LY}=$ $\frac {GE}{GF}\cdot\frac {LX}{LY}\cdot\frac {LF}{LE}.$ Since $\frac {LX}{LY}=\frac {MX}{MY}$ obtain that $\frac {LE}{LF}=\frac {GE}{GF}.$

Observe that $\left\{\begin{array}{ccccccc} \boxed{XB=\frac {ac^2}{a^2+2c^2}} & \implies & XC=BC-XB=a\left(1-\frac {c^2}{a^2+2c^2}\right)=\frac {a\left(a^2+c^2\right)}{a^2+2c^2} & \implies & \boxed{XC=\frac {a\left(a^2+c^2\right)}{a^2+2c^2}} & \implies & \frac {XB}{XC}=\frac {c^2}{a^2+c^2}\\\\ \boxed{YC=\frac {ab^2}{a^2+2b^2}} & \implies & YB=BC-YC=a\left(1-\frac {b^2}{a^2+2b^2}\right)=\frac {a\left(a^2+b^2\right)}{a^2+2b^2} & \implies & \boxed{YB=\frac {a\left(a^2+b^2\right)}{a^2+2b^2}} & \implies & \frac {YC}{YB}=\frac {b^2}{a^2+b^2}\end{array}\right\|\ .$

P10. Let $\triangle ABC$ and $D\in (BC)$ . The incircles of $\triangle ABD$ , $\triangle ACD$ touch $BC$ at $M$ , $N$ respectively.

Prove that $[AD$ is the $A$ -bisector of $\triangle ABC\iff$ $\frac 1{MB}+\frac 1{MD}=\frac 1{ND}+\frac 1{NC}$ .

Proof 1 (Luis Gonzalez). The incircle $(I)$ of $\triangle ABC$ touch $BC$ at $X$ . Then $\frac{1}{BX}+\frac{1}{CX}=\frac{1}{s-b}+\frac{1}{s-c}=\frac{a}{(s-b)(s-c)}=\frac{2 \cot \frac{A}{2}}{h_a}\implies$

$\boxed{\frac{1}{BX}+\frac{1}{CX}=\frac{2 \cot \frac{A}{2}}{h_a}}\ (*)$ because $S=\frac {ah_a}{2}=s(s-a)\tan\frac A2=(s-b)(s-c)\cot\frac A2$ . Using the relation $(*)$ on the triangles $ABD$ and $ACD$

with common $A$ -altitude $h_a$ we get that $\frac{1}{MB}+\frac{1}{MD}=\frac{1}{ND}+\frac{1}{NC} \Longleftrightarrow \cot \frac{\widehat{BAD}}{2}= \cot \frac{\widehat{CAD}}{2} \Longleftrightarrow \widehat{BAD}=\widehat{CAD}\ .$

Proof 2. Denote $\frac {DB}{m}=\frac {DC}{1}=\frac {a}{m+1}$ and $AD=l$ . Using the Stewart's relation obtain that $\boxed{(m+1)^2l^2=(m+1)\left(c^2+mb^2\right)-ma^2}\ (*)$ . Therefore,

$\frac 1{MB}+\frac 1{MD}=\frac 1{ND}+\frac 1{NC}$ $\iff$ $c\cdot ND\cdot NC=$ $b\cdot MB\cdot MD\iff$ $c\cdot\left[DC^2-(l-b)^2\right]=$ $b\cdot\left[DB^2-(l-c)^2\right]$ $\iff$ $c\cdot DC^2-b\cdot DB^2=$

$c(l-b)^2-b(l-c)^2$ $\iff$ $a^2\left(c-m^2b\right)=$ $(m+1)^2(b-c)\left(bc-l^2\right)$ $\stackrel{(*)}{\iff}$ $a^2\left(c-m^2b\right)=$ $bc(b-c)(m+1)^2-(b-c)\left[(m+1)\left(c^2+mb^2\right)-ma^2\right]$

$\iff$ $4(s-b)(s-c)\left[bm^2+(b-c)m-c\right]=0$ $\iff$ $m\in\left\{-1,\frac cb\right\}$ $\iff$ $m=\frac cb\iff$ the ray $[AD$ is the $A$ -bisector of $\triangle ABC$ .

P11. Let $\triangle ABC$ with incircle $w=C(I,r)$ what touches $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ .

Let $\{D,X\}= AD\cap w$ . Prove that $\frac {2}{s-a}=\frac 1{s-b}+\frac 1{s-c}\iff XD=2\cdot XA$ .

Proof 1 (synthetic). Suppose w.l.o.g. $b>c$ . Let $L\in BC\cap EF$ , $T\in AD\cap LI$ - the midpoint of $[XD]$ and $P\in BC\ ,\ AP\perp BC$ . It is well-known (or prove easily) that $LI\perp AD,$ i.e. $ALPT$ is inscribed in the circle with diameter $[AL]\iff$ $\boxed{DP\cdot DL=DT\cdot DA}\ (*)$ . Prove easily that $\boxed{PD=\frac {(b-c)(s-a)}{a}}$ and $\frac {LB}{s-b}=\frac {LC}{s-c}=\frac {a}{b-c}$ $\implies$ $LB=\frac {a(s-b)}{b-c}\implies$ $LD=LB+BD=$ $\frac {a(s-b)}{b-c}+(s-b)\implies$ $\boxed{LD=\frac {2(s-b)(s-c)}{b-c}}$ . In conclusion, $XD=2\cdot XA\iff$ $DT=TX=AX\stackrel{(*)}{\iff}$ $DP\cdot DL=AX\cdot AD\iff$ $DP\cdot DL=AF^2\iff$ $\frac {(b-c)(s-a)}{a}\cdot\frac {2(s-b)(s-c)}{b-c}=(s-a)^2\iff$ $2(s-b)(s-c)=a(s-a)\iff$ $\frac {2}{s-a}=\frac 1{s-b}+\frac 1{s-c}$ .

Proof 2 (metric). Denote $\left\{\begin{array}{c} s-a=x\\\ s-b=y\\\ s-c=z\end{array}\right|\iff$ $\left\{\begin{array}{c} a=y+z\\\ b=z+x\\\ c=x+y\end{array}\right|$ . Apply the Stewart's relation to the cevian $AD$ in $\triangle ABC\ :\ \boxed{a\cdot AD^2+ayz=b^2y+c^2z}\ (*)$ . Thus, $XD=2\cdot XA\iff$ $AX=\frac {AD}{3}\iff$ $\frac {AD}{3}\cdot AD=AF^2\iff$ $AD^2=3x^2\stackrel{(*)}{\iff}$ $3ax^2+ayz=b^2y+c^2z\iff$ $3x^2(y+z)+yz(y+z)=$ $y(x+z)^2+z(x+y)^2\iff$ $3x^2(y+z)+yz(y+z)=x^2(y+z)+4xyz+yz(y+z)\iff$ $x^2(y+z)=2xyz\iff$ $\frac 2x=\frac 1y+\frac 1z$ .

Proof 3 (metric). Denote $\left\{\begin{array}{c} s-a=x\\\ s-b=y\\\ s-c=z\end{array}\right|\iff$ $\left\{\begin{array}{c} a=y+z\\\ b=z+x\\\ c=x+y\end{array}\right|$ . Apply the Pythagoras' relation to the side $[AD]$ in $\triangle ABD\ :\ AD^2=BA^2+BD^2-2\cdot BA\cdot BD\cdot\cos B=$ $(x+y)^2+y^2-2y(x+y)\cos B=$ $x^2+2y(x+y)(1-\cos B)=$ $x^2+4y(x+y)\cdot\sin^2\frac B2=$ $x^2+4y(x+y)\cdot\frac {xz}{(y+x)(y+z)}\implies$ $\boxed{AD^2=x^2+\frac {4xyz}{y+z}}\ (*)$ . In conclusion, $XD=2\cdot XA\iff$ $AX=\frac {AD}{3}\iff$ $\frac {AD}{3}\cdot AD=AF^2\iff$ $AD^2=3x^2\stackrel{(*)}{\iff}$ $x^2+\frac {4xyz}{y+z}=3x^2\iff$ $2yz=x(y+z)\iff$ $\frac 2x=\frac 1y+\frac 1z$ .

Proof 4 (metric). Let $\left\{\begin{array}{c} s-a=x\\\ s-b=y\\\ s-c=z\end{array}\right|\iff$ $\left\{\begin{array}{c} a=y+z\\\ b=z+x\\\ c=x+y\end{array}\right|$ . Pythagoras' relation to $[AD]$ in $\triangle AFD\implies$ $AD^2=FA^2+FD^2-2\cdot FA\cdot FD\cdot\cos \widehat{AFD}=$ $x^2+\left(2y\sin\frac B2\right)^2-2\cdot x\cdot 2y\sin\frac B2\cdot\cos\left(90^{\circ}+\frac B2\right)=$ $x^2+4y(x+y)\cdot\sin^2\frac B2=$ $x^2+4y(x+y)\cdot\frac {xz}{(y+x)(y+z)}\implies$ $\boxed{AD^2=x^2+\frac {4xyz}{y+z}}\ (*)$ . In conclusion, $XD=2\cdot XA\iff$ $AX=\frac {AD}{3}\iff$ $\frac {AD}{3}\cdot AD=AF^2\iff$ $AD^2=3x^2\stackrel{(*)}{\iff}$ $x^2+\frac {4xyz}{y+z}=3x^2\iff$ $2yz=x(y+z)\iff$ $\frac 2x=\frac 1y+\frac 1z$ .

P12. In $\triangle ABC$ Let $D\in AC$ so that $BD\perp AC$ . The tangents at circumcircle $w=C(O)$ of $\triangle ABC$ at $A$

and $C$ intersect at $K$ . Let $F$ of the line $BK$ with $w$ and $E\in BO\cap AC$ . Prove that $\widehat{AFE}\equiv\widehat{CFD}$ .

Proof. Assume $BC \ge AB$ It is well known that $BK$ is $B$ -symmedian of $\Delta ABC$ and also it is $F$ -symmedian of $\Delta ACF$ . Let $M$ be the midpoint of $AC\Longrightarrow$ $\angle MBC =\angle FBA$ . Thus, $\widehat{DBA}\equiv\widehat{EBC} \Longrightarrow$ $\angle MBE=\angle FBD$ . Observe that $\angle FAC=\angle FBC =$ $\angle MBA$ and $\angle AFM =\angle BFC =$ $\angle BAC \Longrightarrow$ $\Delta FAM \sim \Delta ABM \Longrightarrow$ $\angle BMA=\angle FMA$ . Hence we conclude that $D$ and $E$ are isogonally conjugated with respect to $\Delta BMF \Longrightarrow$ $\angle EFM =\angle DFB$ . But $\angle BFC= \angle AFM \Longrightarrow$ $\angle AFE= \angle CFD$ as desired.

P13. Let be given a triangle $ABC$ and its internal angle bisector $BD$ $(D\in BC)$ . The line $BD$ intersects the circumcircle $\Omega$ of $\triangle ABC$ at $B$ and $E$ . Circle $\omega$ with diameter $DE$ cuts $\Omega$ again at $F$ . Prove that $BF$ is the symmedian line of triangle $ABC$ .

Lemma. Let $ABC$ be a triangle with the circumcircle $w$ . Denote the following points : the middlepoint $M$ of the side $[BC]$ ; the point $D\in (BC)$ for which $\angle DAB\equiv\angle DAC$ ; the point $E$ for which $\{A,E\} = AD\cap w$ ; the $A$ -symmedian $AL$ , where $L\in (BC)$ ; the point $S$ for which $\{A,S\} = AL\cap w$ . Then $SD\perp SE$ , i.e. the quadrilateral $MDSE$ is cyclically.

Proof. Let $XX$ - the tangent in $X\in w$ to $w$ and $T\in BB\cap CC$ . Thus, $\overline {MET}\perp BC$ and $\angle ASB\equiv\angle ACB$ . From the first well-known property, $T\in \overline {ALS}$ and the line $\overline{ALST}$ is the $A$ -symmedian in $\triangle ABC$ $\Longrightarrow$ $\left\|\begin{array}{c} \angle BAS\equiv\angle MAC \\ \ \angle SAE\equiv\angle DAM\end{array}\right\|$ . From the second well-known property, the division $\{A,S;L,T\}$ is harmonic. Since $ML\perp MT$ results $\angle DMA\equiv\angle DMS$ . Show easily that $\triangle BAS\sim\triangle MAC$ . Thus, $\frac {AB}{AS} = \frac {AM}{AC}$ , i.e. $AS\cdot AM = AB\cdot AC$ . From the third well-known property $AD\cdot AE = AB\cdot AC$ , obtain $\boxed {AS\cdot AM = AD\cdot AE = AB\cdot AC}$ , i.e. $\frac {AS}{AE} = \frac {AD}{AM}$ . Since $\angle SAE\equiv\angle DAM$ obtain $\triangle SAE\sim\triangle DAM$ and $\angle DMA\equiv\angle SEA$ . Since $\angle DMA\equiv\angle DMS$ and $\angle SEA\equiv\angle SED$ obtain $\angle DMS\equiv\angle SED$ , i.e. $MDSE$ is cyclic. In conclusion, $SD\perp SE$ .

Remark. Interesting relations : $\left\|\begin{array}{ccc} ABS\sim CMS & \implies & \boxed {SB\cdot SC = SA\cdot SM} \\ \\ AMC\sim CMS & \implies & \boxed {MA\cdot MS = MB^2}\end{array}\right\|$ . PP13 is a consequence of the above lemma. We can use it to PP from here

P14. A semicircle $h(O)$ has diameter $[XY]\subset [BC]\subset\triangle ABC$ and it is tangent to $(AB)$ and $(AC)$ in $F$ and $E.$ Let $G\in XE\cap YF$ . Prove that $AG\perp BC$

Proof. Suppose w.l.o.g. $b<c$ and $X\in (OB)$ . Denote $\left\{\begin{array}{c} H\in EX\cap AO\\\ K\in EX\cap AO\end{array}\right|$ . Observe that $AO$ is the $A$ -bisector of $\triangle ABC$ , i.e. $KE=KF$ , $HE=HF$ , $OX=OF=OE=OY$ . Therefore, $\left\{\begin{array}{ccc} \left|\begin{array}{c} m\left(\widehat{BOF}\right)=90^{\circ}-B\\\\ OF=OY\end{array}\right| & \implies & m\left(\widehat{XYF}\right)=45^{\circ}-\frac B2\\\\ \left|\begin{array}{c} m\left(\widehat{COE}\right)=90^{\circ}-C\\\\ OE=OX\end{array}\right| & \implies & m\left(\widehat{YXE}\right)=45^{\circ}-\frac C2\end{array}\right\|\implies$ $m\left(\widehat{EGY}\right)=m\left(\widehat{XYF}\right)+m\left(\widehat{YXE}\right)=\frac A2\implies$ $m\left(\widehat{EGY}\right)=\frac A2$ . In conclusion, $AFGH$ and $EFXY$ are cyclically $\implies$ $m\left(\widehat{XEF}\right)=m\left(\widehat{XYF}\right)=45^{\circ}-\frac B2$ . Since $HE=HF$ obtain that $m\left(\widehat{XHF}\right)=2\cdot m\left(\widehat{XEF}\right)=90^{\circ}-B$ . Thus, $\widehat{XHF}\equiv\widehat{BAG}\implies$ $m\left(\widehat{BAG}\right)=90^{\circ}-B\implies$ $AG\perp BC$ .

P15. Denote the circumcircle $\Gamma = \mathcal C(O,R)$ and the incircle $w = \mathcal C(I,r)$ of a right triangle $ABC$ , where $AB\perp AC$ . Denote : the midpoints $M$ , $N$ of the sides $[AB]$ , $[AC]$ ; $\{P,Q\} = MN\cap \Gamma$ so that $M\in (PN)$ , $N\in (MQ)$ ; $D\in (AB)\cap w$ , $E\in (AC)\cap w$ . Prove that the points $D$ , $E$ , $P$ , $Q$ are concyclic.

Proof (metric). The bisectors of the segments $[DE]$ , $[PQ]$ meet $\Gamma$ in the same point $A'$ . Then $D$ , $E$ , $P$ , $Q$ are concyclic $\iff \boxed {A'P = A'D}$ and in this case these points belong to the circle with center $A'$ . I'll ascertain the lengths $A'P$ and $A'D$ ! Let the length $h$ of the $A$ - altitude of $\triangle ABC$ .

The first method (use only the theorem of cathetus and the Pythagoras' theorem).

$1\blacktriangleright$ Let diameter $[A'A_1]$ of $\Gamma$ and $R\in A'A_1$ , $PR\perp A'A_1$ . Thus, $A'P^2 = A'A_1\cdot A'R =$ $a\left(\frac a2 + \frac h2\right) =$ $a\left(\frac a2 + \frac {bc}{2a}\right)$ , i.e. $\boxed {A'P = \sqrt {\frac {a^2 + bc}{2}}}\ .$

$2\blacktriangleright$ Let $B_1$ for which $BC$ doesn't separate $B_1$ , $A$ and $B_1B\perp BC$ , $B_1B = r$ . Thus, $\triangle B_1BA'\equiv\triangle DIA'$ $\implies$ $\boxed {A'B_1 = A'D}$ .

Prove easily $A'B_1^2 = \left(\frac a2\right)^2 + \left(\frac a2 + r\right)^^2$ . Thus, $r = p - a$ $\implies$ $A'B_1^2 = \frac 14\cdot\left[a^2 + (b + c)^2\right]$ $\implies$ $A'B_1 = \sqrt {\frac {a^2 + bc}{2}}$ $\implies$ $\boxed {A'D = \sqrt {\frac {a^2 + bc}{2}}}\ .$

$\blacksquare$ In conclusion, $A'P = A'D$ , i.e. $\{\ D\ ,\ E\ ,\ P\ ,\ Q\ ,\ B_1\ \}\subset \mathcal C\left(A'\right)\ .$

The second method (use the Ptolemaeus' and the generalized Pythagoras' theorems).

$1\blacktriangleright$ $\left\|\begin{array}{c} PB^2 + PC^2 = a^2 \\ \\ PB\cdot PC=\frac {bc}2\end{array}\right\|$ $\implies$ $PB + PC = \sqrt {a^2 + bc}$ . Ptolemy' theorem to $BPCA':\ A'P\cdot BC = BA'\cdot (PB + PC)$ $\implies$ $\boxed {A'P = \sqrt {\frac {a^2 + bc}{2}}}$

$2\blacktriangleright$ The Ptolemy' theorem to $BACA'\ :\ AA' = \frac {\sqrt 2}{2}\cdot (b + c)$ . The generalized Pythagoras' theorem $:\ A'D^2 = AD^2 + A'A^2 - \sqrt 2\cdot AD\cdot AA' =$

$(p - a)^2 + \frac 12\cdot (b + c)^2 - (p - a)(b + c)\ .$ Thus, $A'D^2 = \frac 12\cdot (b + c)^2 - p(p - a) =$ $\frac 12\cdot (b + c)^2 - \frac {bc}{2}\implies$ $\boxed {A'D = \sqrt {\frac {a^2 + bc}{2}}}$ .

$\blacksquare$ In conclusion, $A'P = A'D$ , i.e. $\{\ D\ ,\ E\ ,\ P\ ,\ Q\ \}\subset \mathcal C\left(A'\right)\ .$

P16. An acute triangle $ABC$ with $AB \neq AC$ is given. Let $V$ and $D$ be the feet of the altitude and angle bisector from $A$ , and let $E$ and $F$

be the intersection points of the circumcircle of $\triangle AVD$ with sides $AC$ and $AB$ , respectively. Prove that $AV$ , $BE$ and $CF$ have a common point.

An easy extension. Let $ABC$ be an acute triangle with $c < b$ and let $M\in (BC)$ , $N\in (MC)$ so that $\left\{\begin{array}{c} x = m(\widehat {AMC}) \\ \ y = m(\widehat {ANB})\end{array}\right\|$ .The circumcircle of $\triangle AMN$ cut

again the sides $[AB]$ , $[AC]$ in the points $F$ , $E$ respectively. Then the lines $BE$ , $CF$ , $AM$ are concurrently $\Longleftrightarrow$ $\cos (C - B + x - 2y) + \cos A\cos x = 0$ .

Particular cases.

$1\blacktriangleright\ x: = 90^{\circ}$ , i.e. $AM\perp BC$ . In this case, $\left\{\begin{array}{c} C - B + 90^{\circ} - 2y = \pm 90^{\circ} \\ \ (\ C\ < \ B\ )\end{array}\right\|$ $\Longleftrightarrow$ $y = C + \frac A2$ $\Longleftrightarrow$ $\widehat {NAB}\equiv\widehat {NAC}$ .

$2\blacktriangleright\ A: = 90^{\circ}$ . In this case obtain the following nice problem :

P17. Let $ABC$ be a $A$ - right triangle, i.e. $AB\perp AC$ with $c < b$ . Choose two points $\left\{\begin{array}{c} M\in (BC) \\ \ N\in (MC)\end{array}\right\|$ . The circumcircle $w$ of the triangle $AMN$

cut again the sides $[AB]$ , $[AC]$ in the points $F$ , $E$ respectively. Then the lines $BE$ , $CF$ , $AM$ are concurently $\Longleftrightarrow$ $m(\widehat {AMC}) = 2\cdot m(\widehat {ANC})$ .

$3\blacktriangleright$ Here is a very easy problem which, for $A = 90^{\circ}$ , is equivalently with the previous problem :

P18. Let $ABC$ be a triangle. For $M\in (BC)$ denote $\left\{\begin{array}{c} F\in (AB)\ ,\ \widehat {FMA}\equiv\widehat {FMB} \\ \\ E\in (AC)\ ,\ \widehat {EMA}\equiv\widehat {EMC}\end{array}\right\|$ . Then the lines $BE$ , $CF$ , $AM$ are concurently.

Extension. Let $\triangle ABC,$ $M\in (BC)$ and $\left\{\begin{array}{ccccc} F\in (AB) & : & m\left(\widehat {FMB}\right)=\alpha & ; & m\left(\widehat {FMA}\right)=\beta\\\\ E\in (AC) & : & m\left(\widehat {EMA}\right)=\gamma & ; & m\left(\widehat{EMC}\right)=\delta\end{array}\right\|\ .$ Prove that $BE\cap CF\cap AM\ne \emptyset \iff \sin\alpha\sin\gamma=\sin\beta\sin \delta .$

P19 (M.O. Sanchez). Let $\triangle ABC$ with centroid $G$ and its cevian $\triangle DEF,$ where $D\in BC,$ $E\in CA,$ $F\in AB.$ Prove $\odot\begin{array}{cc} \nearrow AEGF\ \mathrm{is\ cyclic}\iff BC=AG\sqrt 3\\\\ \searrow AEGF\ \mathrm{is\ tangential}\iff AB=AC\end{array}$

Proof. $a=k\cdot AG\iff$ $3a=2km_a\iff$ $9a^2=k^2\left[2\left(b^2+c^2\right)-a^2\right]\iff$ $\boxed{a^2\left(k^2+9\right)=2k^2\left(b^2+c^2\right)}\ (*)$ and $AEGD$ is cyclic $\iff$

$BF\cdot BA=BG\cdot BE\iff$ $\frac c2\cdot c=\frac {2m_b}3\cdot m_b\iff$ $3c^2=4m_b^2\iff$ $3c^2=2\left(a^2+c^2\right)-b^2\iff$ $b^2+c^2=2a^2\ \stackrel{(*)}{\iff}$

$4a^2k^2=a^2\left(k^2+9\right)\iff$ $4k^2=k^2+9\iff$ $k=\sqrt 3\iff$ $a=AG\sqrt 3.$ Apply property from proposed problem PP1 $:\ AEGF$ is circumscriptibly $\iff$

$\left\{\begin{array}{ccccccc} GB-GC & = & AB-AC & \iff & \frac 23\cdot\left(m_b-m_c\right) & = & c-b\\\\ EB-FC & = & FB-EC & \iff & m_b-m_c & = & \frac 12\cdot (c-b)\end{array}\right\|$ $\iff$ $AB=AC.$

P20 (M. O. Sanchez). Let $\triangle ABC$ with centroid $G$ and midpoints $D,$ $F$ of $[BC],$ $[BA]$ respectively.

Suppose that $C>90^{\circ}$ and $BDGF$ is circumscriptibly. Prove that $BC\sqrt 2<AC.$

Proof. Observe that $C>90^{\circ}\iff \boxed{a^2+b^2<c^2}\ (1).$ Therefore, $BDGF$ is tangential $\iff$ $CD\cdot CB=CG\cdot CF\iff$ $\frac a2\cdot a=\frac {2m_c}3\cdot m_c\iff$ $3a^2=4m_c^2\iff$

$3a^2=2\left(a^2+b^2\right)-c^2\iff$ $\boxed{a^2+c^2=2b^2}\ (2)\ .$ The sum of $(1)$ and $(2)\ \implies\$ $\left(\underline a^2+\cancel{b^2}\right)+\left(\underline a^2+\cancel{c^2}\right)<\cancel{c^2}+\cancel 2b^2\implies$ $2a^2<b^2\implies a\sqrt 2<b$ $\implies BC\sqrt 2<AC.$

P21 (M. O. Sanchez). Let $\triangle ABC$ with the circumcircle $w=\mathbb C(O,R).$ Denote $:$ the diameter $[PQ]$ of $w$ so that $PQ\perp BC,$ $D\in PQ\cap BC$

and $BC$ separates $Q$ and $A\ ;$ the projections $M$ and $N$ of $P$ and $Q$ respectively on $AC.$ Prove that $MN=AB$ and $DM\perp DN.$

Proof. Suppose w.l.o.g. $B>C$ and let $X\in PQ\cap AC.$ Hence $MN=PQ\cos\widehat {PXA}=2R\cos\left(90^{\circ}-C\right)=$ $2R\sin C=$ $AB\implies$ $\boxed{MN=AB}\ .$

$\left\{\begin{array}{ccc} \mathrm{DMPC\ is\ cyclic}\iff m\left(\widehat{MDP}\right)=m\left(\widehat{MCP}\right) & = & \frac {B-C}2\\\\ \mathrm{DQCN\ is\ cyclic}\iff m\left(\widehat{NDC}\right)=m\left(\widehat{NQC}\right) & = & 90^{\circ}-\left(C+\frac A2\right)\end{array}\right\|\ \stackrel{\frac {B-C}2+C+\frac A2=90^{\circ}}{\implies}\ \widehat {MDP}\equiv\widehat{NDC}\implies$ $\widehat {MDN}\equiv\widehat {PDC}\implies$ $\boxed{\ DM\perp DN}\ .$

P22 (M. O. Sanchez). Let $ABCDEFG$ be a regular heptagon for which denote $\left\{\begin{array}{ccc} AB & = & a\\\ AC & = & b\\\ AD & = & c\end{array}\right\|\ .$ Prove that $\frac 1a=\frac 1b+\frac 1c$ and $\frac {c^3}{b^3}=\frac {b+2c}{a+c}\ .$

Proof. I"ll apply the Ptolemy's theorem to the following cyclic quadrilaterals $:\ \left\{\begin{array}{ccccc} ABCD\ : & b^2 & = & a(a+c) & (1)\\\ ABCE\ : & bc & = & a(b+c) & (2)\\\ ABDE\ : & c^2 & = & a^2+bc & (3)\\\ ABDF\ : & c^2 & = & b(a+b) & (4)\end{array}\right\|\ .$ From the relation $(2)$ obtain

that $\frac 1a=\frac 1b+\frac 1c.$ Therefore, $\frac {c^3}{b^3}=\frac {c\cdot c^2}{b\cdot b^2}\ \stackrel {1\wedge 4}{=}\ \frac {c\cdot b(a+b)}{b\cdot a(a+c)}=$ $\frac {c(a+b)}{a(a+c)}=$ $\frac {bc+ac}{a(a+c)}\ \stackrel {2}{=}\ \frac {a(b+c)+ac}{a(a+c)}=$ $\frac {(b+c)+c}{a+c}=$ $\frac {b+2c}{a+c}$ $\implies$ $\frac {c^3}{b^3}=\frac {b+2c}{a+c}\ .$

Remark. Observe that we didn't use the relation $(3).$ Can obtain easily the relation $(3)$ from the relations $(1),$ $(2)$ and $(4).$

Indeed, $c^2\ \stackrel{4}{=}\ b(a+b)=$ $ab+b^2\ \stackrel{1}{=}\ ab+a(a+c)=$ $a^2+a(b+c)\ \stackrel{2}{=}\ a^2+bc$ $\implies$ $c^2=a^2+bc\ .$

P23 (France TST 2007). Let $\triangle ABC$ such that $C<A<\frac{\pi}{2}$ . Let $D\in [AC]$ such that $BD=BA$ . The incircle $w=C(I,r)$ of $\triangle ABC$ touches

its sides in $K\in [AB]$ and $L\in [AC]$ . Let $w_{1}=C(J,r_{1})$ be the incircle of $\triangle BCD$ . Denote $T\in KL\cap AJ$ . Prove that $TA=TJ$ .

Proof.. The incircle $w_{1}$ touches $\triangle BCD$ in $U\in BD$ , $V\in BC$ and $W\in BC$ . Observe that $DC=AC-AD=b-2c\cdot\cos A=$ $\frac{b^{2}-(b^{2}+c^{2}-a^{2})}{b}$ $\implies$

$DC=\frac{a^{2}-c^{2}}{b}$ . Thus, $VC=\frac{1}{2}\cdot (BC+CD-BD)=\frac{1}{2}\cdot\left(a+\frac{a^{2}-c^{2}}{b}-c\right)\implies VC=\frac{p(a-c)}{b}$ . Since $VC=JC\cdot\cos \frac{C}{2}$ obtain that $\boxed{JC=\frac{p(a-c)}{b\cdot\cos\frac{C}2}}$ .

Denote $P\in KL\cap CI$ . The property $PC\perp PB$ is well-known (see the lower remark). Thus, $APJW$ is cyclically. Analogously $P\in UV\cap CJ$ , i.e. $\boxed{\ P\in KL\cap UV\ }$ .

Therefore, $\boxed{PC=a\cdot \cos \frac{C}{2}}\implies$ $\frac{PC}{JC}=\frac{ab\cdot\cos^{2}\frac{C}{2}}{p(a-c)}=\frac{p-c}{a-c}$ $\implies$ $\frac{PC}{p-c}=\frac{JC}{a-c}=\frac{PJ}{p-a}$ $\implies$ $\boxed{\ \frac{PJ}{PC}=\frac{p-a}{p-c}\ }$ . Apply the Menelaus' theorem to the transversal

$\overline{PTL}$ in $\triangle AJC\ :\ \frac{PJ}{PC}\cdot\frac{LC}{LA}\cdot\frac{TA}{TJ}=1$ $\implies$ $\frac{p-a}{p-c}\cdot\frac{p-c}{p-a}\cdot\frac{TA}{TJ}=1$ $\implies TA=TJ$ . Thus, $PC\perp PB$ $\Longleftrightarrow$ the quadrilateral $PKIB$ is cyclic $\Longleftrightarrow$

$m(\widehat{PKB})=m(\widehat{PIB})=\frac{B+C}{2}$ or $m(\widehat{PKA})=m(\widehat{PIB})=\frac{B+C}{2}$

Lemma. Let $ABC$ be a triangle with $E\in (AC),$ $F\in (AB),$ $G\in (EF)$ and $M\in AG\cap BC.$ Then there is the relation $\frac {MB}{MC}=\frac {GF}{GE}\cdot\frac {AB}{AF}\cdot \frac {AE}{AC}.$

Proof. Denote $L\in EF\cap BC$ and suppose w.l.o.g. $B\in LC.$ Apply the Menelaus' ttheorem to the transversals $:$

$\left\{\begin{array}{ccccc} \overline{AGM}/\triangle FBL\ : & \frac {GF}{GL}\cdot\frac {ML}{MB}\cdot\frac {AB}{AF}=1\\\\ \overline{AGM}/\triangle CEL\ : & \frac {AE}{AC}\cdot\frac {MC}{ML}\cdot \frac {GL}{GE}=1\end{array}\right\|\ \bigodot\implies$ $\frac {GF}{\cancel {GL}}\cdot\frac {\cancel{ML}}{MB}\cdot\frac {AB}{AF}\cdot\frac {AE}{AC}\cdot\frac {MC}{\cancel{ML}}\cdot \frac {\cancel{GL}}{GE}=1\iff$ $\frac {MB}{MC}=\frac {GF}{GE}\cdot\frac {AB}{AF}\cdot \frac {AE}{AC}.$

P24 (the cevian nest theorem). Given $\triangle ABC$ with $\left\{\begin{array}{ccc} D\in (BC)\ ,\ E\in (CA)\ ,\ F\in (AB) & : & AD\cap BE\cap CF\ne\emptyset\\\\ G\in (EF)\ ,\ H\in (FD)\ ,\ K\in (DE) & : & DG\cap EH\cap FK\ne \emptyset\end{array}\right\|\ .$ Prove that $AG\cap BH\cap CK\ne\emptyset .$

Proof. Denote $\left\{\begin{array}{ccccc} \frac {DB}x=\frac {DC}1=\frac {BC}{x+1} & ; & M\in AG\cap BC\\\ \frac {EC}y=\frac {EA}1=\frac {CA}{y+1} & ; & N\in BH\cap CA\\\ \frac {FA}z=\frac {FB}1=\frac {AB}{z+1} & ; & P\in CK\cap AB\end{array}\right\|$ where $xyz=1,$ i.e. $AD\cap BE\cap CF\ne\emptyset .$ Observe that $\frac {GE}{GF}\cdot \frac {HF}{HD}\cdot\frac {KD}{KE}=1.$ Apply upper lemma $:$

$\left\{\begin{array}{ccc} \frac {MB}{MC}=\frac {GF}{GE}\cdot\frac {AB}{AF}\cdot \frac {AE}{AC} & \iff & \frac {MB}{MC}=\frac {GF}{GE}\cdot \frac {z+1}{z(y+1)}\\\\ \frac {NC}{NA}=\frac {HD}{HF}\cdot\frac {BC}{BD}\cdot \frac {BF}{BA} & \iff & \frac {NC}{NA}=\frac {HD}{HF}\cdot \frac {x+1}{x(z+1)}\\\\ \frac {PA}{PB}=\frac {KE}{KD}\cdot\frac {CA}{CE}\cdot \frac {CD}{CB} & \iff & \frac {PA}{PB}=\frac {KE}{KD}\cdot\frac {y+1}{y(x+1)}\end{array}\right\|\ \bigodot\implies$ $\frac {MB}{MC}\cdot\frac {NC}{NA}\cdot\frac {PA}{PB}=1\implies$ $AM\cap BN\cap CP\ne\emptyset\implies$ $AG\cap BH\cap CK\ne\emptyset .$

P25. Let $C(O)$ , $C(T)$ be the circumcircles of an acute $\triangle ABC$ and $\triangle AOC$ respectively. Let $M$ be the midpoint of $[AC].$

Consider the points $D\in [AB]$ and $E\in [BC]$ which satisfy $\widehat {BDM}\equiv\widehat {BEM}\equiv\widehat {ABC}$ . Show that $BT\perp DE$ .

Proof. Let $\left\{\begin{array}{c} A'\in BA\ ,\ A'B = A'C \\ \\ C'\in BC\ ,\ C'B = C'A\end{array}\right\|$ . Observe that $m(\widehat {AOC}) = 2B$ and $m(\widehat {AA'C}) = m(\widehat {AC'C}) = 180^{\circ} - 2B$ , i.e. $A'$ , $C'$ belong to the circle $C(T)$ . Denote the midpoints

$X$ , $Y$ of the segments $[AA']$ , $[CC']$ respectively. Observe that $\left\{\begin{array}{c} MX\parallel A'C\implies E\in XM \\ \\ MY\parallel C'A\implies D\in MY\end{array}\right\|$ and $\left \{\begin{array}{c} TX\perp BA \\ \\ TY\perp BC\end{array}\right\|\implies$ the quadrilateral $BXTY$ is ciclically (the circumcenter

is the midpoint of $[BT]$ ). From the relations $m(\widehat {XDY}) = m(\widehat {XEY}) = 180^{\circ} - B$ obtain that the quadrilateral $DEYX$ is cyclically. Thus, the perpendicular line $d$ from the point $B$

to the line $DE$ passes through the center of the circumcircle of $\triangle BXY$ (and of the quadrilateral $BXTY$ ) , i.e. the midpoint $[BT]$ . Thus, $d\equiv BT$ . In conclusion, $BT\perp DE$ .

This post has been edited 377 times. Last edited by Virgil Nicula, Dec 12, 2016, 10:22 AM

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358. Some properties of the remarkable lines in a triangle.

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