146. Synthetic, metric, trigonometric and analitycal proofs.

by Virgil Nicula, Oct 5, 2010, 5:44 PM

Let $ABC$ be an $A$-isosceles triangle. Denote the midpoint $D$ of $[BC]$ , the foot $E$ of the perpendicular from $D$ to $AB$ and the midpoint $F$ of $[DE]$ . Prove that $AF\perp CE$ .

Proof 1 (synthetic). Let $I\in CE\cap AF$ and projection $G$ of $C$ to $AB$. Thus, $DE\parallel CG$ and $EG=EB$, i.e. ray $[CE$ is $C$- median in $\triangle CGB$. Prove easily that $ADE\sim CBG$.

$A$-median of $\triangle ADE$ is similarly with $C$- median of $\triangle CGB$ and $[AD]$ of $\triangle ADE$ is similarly with $[CB]$ of $\triangle CBG$. Thus, $\widehat{DAF}\equiv\widehat{BCE}$, i.e. $\widehat{DAI}\equiv\widehat{DCI}$, what means that $ACDI$

is cyclic, i.e. $\widehat{ADC}\equiv\widehat{AIC}$. But $DA\perp DC$. In conclusion, and $IA\perp IC$, i.e. $AF\perp CE\ .$

Proof 2 (metric). $AC^{2}-AE^{2}=AC^{2}-(AD^{2}-DE^{2})=$ $(AC^{2}-AD^{2})+DE^{2}=DC^{2}+DE^{2}$ $\Longrightarrow$ $\boxed{\ AC^{2}-AE^{2}=DC^{2}+DE^{2}\ }\ \ (1)\ .$ $4\cdot (FC^{2}-FE^{2})=$

$2\cdot (CE^{2}+CD^{2})-DE^{2}-DE^{2}=$ $2\cdot (CE^{2}+DB^{2}-DE^{2})=$ $2\cdot (CE^{2}+BE^{2})=$ $4\cdot DE^{2}+CB^{2}=$ $4\cdot (DC^{2}+DE^{2})$ $\Longrightarrow$ $\boxed{\ FC^{2}-FE^{2}=DC^{2}+DE^{2}\ }\ \ (2)\ .$

From $(1)$ and $(2)$ get $AC^{2}-AE^{2}=FC^{2}-FE^{2}$, i.e. $\boxed{\ AF\perp CE\ }\ .$

Proof 3 (trigonometric). Let $\phi =m(\widehat{DAF})$ and $\psi =m(\widehat{BCE})$. Thus, $1=\frac{FD}{FE}=$ $\frac{AD}{AE}\cdot\frac{\sin\phi}{\sin\left(\frac{A}{2}-\phi\right)}=$ $\frac{\sin\phi}{\cos\frac{A}{2}\sin\left(\frac{A}{2}-\phi\right)}$ $\Longrightarrow$ $2\sin\phi =\sin (A-\phi )-\sin \phi$ $\Longrightarrow$

$3\sin \phi =\sin (A-\phi )$ $\Longrightarrow$ $3\tan\phi =\sin A-\cos A\tan\phi$ $\Longrightarrow$ $\boxed{\ \tan\phi =\frac{\sin A}{3+\cos A}\ }\ \ (1)\ .$ $\frac{EB}{EA}=\left(\frac{DB}{DA}\right)^{2}=$ $\tan^{2}\frac{A}{2}\ ,\ \frac{EB}{EA}=$ $\frac{CB}{CA}\cdot\frac{\sin\psi }{\sin \left(90-\frac{A}{2}-\psi \right)}=$

$\frac{2\sin\frac{A}{2}\sin\psi }{\cos\left(\frac{A}{2}+\psi\right)}\ .$ Therefore, $\sin\frac{A}{2}\cos\left(\frac{A}{2}+\psi \right)=$ $2\cos^{2}\frac{A}{2}\sin\psi$ $\Longrightarrow$ $\sin (A+\psi )-\sin \psi =$ $2\sin\psi (1+\cos A)$ $\Longrightarrow$ $\sin A\cos\psi=3\sin\psi+\sin\psi \cos A$ $\Longrightarrow$

$\sin A=\cos A\tan \psi+3\tan\psi$ $\Longrightarrow$ $\boxed{\ \tan\psi =\frac{\sin A}{3+\cos A}\ }\ \ (2)\ .$ From the relations $(1)$ and $(2)$ obtain $\phi =\psi$ $\Longrightarrow$ the quadrilateral $ACDI$ is cyclically $\Longrightarrow$ $AF\perp CE\ .$

Proof 4 (analytic). $D(0,0)\ ,\ B(0,b)\ ,\ C(-b,0)\ ,\ A(0,a)$, where $a>0$, $b>0$. The slope of $AB$ is $s(AB)=-\frac{a}{b}$, the slope of $DE$ is $s(DE)=-\frac{1}{s(AB)}=\frac{b}{a}$ and $E$ is the

intersection between $AB$ with equation $\frac{x}{b}+\frac{y}{a}=1$ and $DE$ with equation $y=\frac{b}{a}x$. Thus, $E$ has coordinates $x_{E}=\frac{a^{2}b}{a^{2}+b^{2}}$ and $y_{E}=\frac{b^{2}a}{a^{2}+b^{2}}$. The middlepoint $F$ of $[DE]$ has

coordinates $x_{F}=\frac{1}{2}\cdot \frac{a^{2}b}{a^{2}+b^{2}}$ and $y_{F}=\frac{1}{2}\cdot \frac{b^{2}a}{a^{2}+b^{2}}$. Thus, the slope of $CE$ is $s(CE)=\frac{\frac{b^{2}a}{a^{2}+b^{2}}}{\frac{a^{2}b}{a^{2}+b^{2}}+b}$ $\iff$ $\boxed{s(CE)=\frac{ab}{2a^{2}+b^{2}}}$ and the slope of $AF$ is $s(AF)=\frac{\frac{b^{2}a}{2(a^{2}+b^{2})}-a}{\frac{a^{2}b}{2(a^{2}+b^{2})}}$

$\iff$ $\boxed{s(AF)=-\frac{2a^{2}+b^{2}}{ab}}$ . In conclusion, $s(CE)\cdot s(AF)=-1$ $\iff$ $CE\perp AF$ .

Remark. I'll prove similarly the following "two problems of the butterfly" :


$\blacksquare\ P.B.1$ Given are a circle $w(O)$, a line $l$ and the point $F\in l$. Let $\{M,N,P,R\}\subset w$ be four points so that

$F\in MN\cap PR$. Denote the points $A\in MR\cap l$, $B\in NP\cap l$. Prove that $\boxed{\ OF\perp l\Longleftrightarrow FA=FB\ }\ .$


Proof. Denote the projections $U$, $V$ of the center $O$ to the lines $MR$, $PN$ respectively. Thus, $\triangle FNP\sim\triangle FRM$ $\Longrightarrow$ $\widehat{FUM}\equiv\widehat{FVP}$

because the sides $[FM]$, $[RM]$ of the triangle $FRM$ are similarly with the sides $[FP]$, $[NP]$ of the triangle $FNP$ $\Longrightarrow$ $\widehat{FUA}\equiv\widehat{FVB}\ \ (1)$.

$\triangleright$ Suppose that $OF\perp l$ . The quadrilaterals $AFUO$, $BFVO$ are cycclically with the diameters $[AO]$, $[BO]$ respectively $\Longrightarrow$ $\widehat{FUA}\equiv\widehat{FOA}$

and $\widehat{FVB}\equiv\widehat{FOB}$. From the relation $(1)$ obtain $\widehat{FOA}\equiv\widehat{FOB}$. From $OF\perp l$ results that the triangle $AOB$ is isosceles, i.e. $FA=FB$ .

$\triangleleft$ Suppose that $FA=FB\ .$ Denote the intersections $X$, $Y$ of the bisector line of the segment $[AB]$ with the lines $OU$, $OV$ respectively. Since

$AFXU$, $BFYV$ are cyclically obtain $\widehat{FXA}\equiv\widehat{FUA}$ , $\widehat{FYB}\equiv\widehat{FVB}$. From $(1)$ obtain $\widehat{FXA}\equiv\widehat{FYB}$, i.e. $X\equiv Y\equiv O$ $\Longrightarrow$ $OF\perp l$ .


$\blacksquare\ P.B.2$ Given are the circle $w(O)$, the points $\{A,B,C,D\}\subset w$ and a line $l$ for which $O\in l$ . Denote :

$M_{1}\in AB\cap l$ ; $M_{2}\in AD\cap l$ ; $M_{3}\in AC\cap l$ ; $N_{1}\in CD\cap l$ ; $N_{2}\in BC\cap l$ ; $N_{3}\in BD\cap l$.

Prove that the following chain - equivalence : $\boxed{\ OM_{1}=ON_{1}\Longleftrightarrow OM_{2}=ON_{2}\Longleftrightarrow OM_{3}=ON_{3}\ }\ .$


Proof (metric). Denote the intersection $R\in AC\cap BD$. The powers of the points $R$ , $N_{3}$ , $M_{3}$ w.r.t. $w$ are :

$p_{w}(R)=RA\cdot RC=RB\cdot RD$ : $p_{w}(N_{3})=N_{3}D\cdot N_{3}B$ ; $p_{w}(M_{3})=M_{3}A\cdot M_{3}C$.

$\blacktriangleright$ Apply the Menelaus' theorem to the transversals $\overline{AM_{2}D}/M_{3}RN_{3}$ and $\overline{BN_{2}C}/M_{3}RN_{3}\ :\ \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot \frac{DN_{3}}{DR}\cdot\frac{AR}{AM_{3}}=1$

and $\frac{N_{2}N_{3}}{N_{2}M_{3}}\cdot\frac{CM_{3}}{CR}\cdot\frac{BR}{BN_{3}}=1$ $\Longrightarrow$ $\frac{N_{2}M_{3}}{N_{2}N_{3}}\cdot \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac{RA}{RD}\cdot\frac{RC}{RB}\cdot\frac{N_{3}D}{M_{3}C}\cdot\frac{N_{3}B}{M_{3}A}=1$, i.e. $\boxed{\ \frac{M_{2}M_{3}}{M_{2}N_{3}}\cdot\frac{N_{2}M_{3}}{N_{2}N_{3}}=\frac{p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (1)\ .$

$\blacktriangleright$ Apply the Menelaus' theorem to the transversals $\overline{M_{1}AB}/M_{3}RN_{3}$ and $\overline{N_{1}CD}/M_{3}RN_{3}\ :\ \frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{BN_{3}}{BR}\cdot\frac{AR}{AM_{3}}=1$

and $\frac{N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac{DN_{3}}{DR}\cdot\frac{CR}{CM_{3}}=1$ $\Longrightarrow$ $\frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{N_{1}M_{3}}{N_{1}N_{3}}\cdot\frac{RA}{RB}\cdot\frac{RC}{RD}\cdot\frac{N_{3}B}{M_{3}A}\cdot\frac{N_{3}D}{M_{3}C}=1$, i.e. $\boxed{\ \frac{M_{1}M_{3}}{M_{1}N_{3}}\cdot\frac{N_{1}M_{3}}{N_{1}N_{3}}=\frac{p_{w}(M_{3})}{p_{w}(N_{3})}\ }\ \ (2)\ .$

From the relations $(1)$, $(2)$ obtain the conclusion of the our problem.

A very interesting particular case, when $A\equiv B$ .


Let $ABC$ be an acute triangle inscribed in the circle $w=C(O,R)$ . Denote the point $A'\in w\cap (AO$

and $P\in BC\cap A'A'$, $M\in AB\cap PO$, $N\in AC\cap PO$. Then exists the relation $OM=ON\ .$
This post has been edited 24 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:02 AM

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