230. Some usual synthetical properties in a triangle.

by Virgil Nicula, Feb 24, 2011, 1:58 PM

PP1. Let $\triangle ABC$ with incircle $w=C(I)$ . Note $E\in AC\cap w$ , $F\in AB\cap w$ , $P\in EF\cap BI$ . Prove that $PB\perp PC$ .

Proof 1. $m\left(\widehat {IBC}\right)=m\left(\widehat {FBP}\right)=$ $\frac B2\ \wedge\ m\left(\widehat {BIC}\right)=$ $m\left(\widehat {BFP}\right)=90^{\circ}+\frac A2$ $\Longrightarrow$ $\triangle BIC\ \stackrel{(A.A)}{\sim}\  \triangle BFP$ $\Longrightarrow$

$\frac{BF}{BI}=$ $\frac{BP}{BC}\ \wedge\ m\left(\widehat {IBC}\right)=$ $m\left(\widehat {FBP}\right)$ $\Longrightarrow$ $\triangle BFI \stackrel{(S.A.S)}{\sim} \triangle BPC$ . In conclusion, $FB\perp FI\implies PB\perp PC$ .

Proof 2. $m\left(\widehat {PEC}\right)=m\left(\widehat {PIC}\right)=$ $90^{\circ}-\frac A2\Longrightarrow PEIC$ is cyclically $\Longrightarrow$ $m\left(\widehat {IPC}\right)=m\left(\widehat {IEC}\right)=90^{\circ}\Longrightarrow PB\perp PC$ .



PP2. Let $\triangle ABC$ with incircle $w=C(I,r)$ . Note midpoint $M$ of $[BC]$ and $R\in MI$ for which $AR\perp BC$ . Prove that $AR=r$ .

Proof. Note $D\in AR\cap MI$ and $L\in AI\cap BC$ . Apply the Menelaus' theorem to the transversal $\overline{MIR}/\triangle ADL\ :$

$\frac {ML}{MD}\cdot\frac {RD}{RA}\cdot\frac {IA}{IL}=1$ $\Longleftrightarrow$ $\frac{\frac {a|b-c|}{2(b+c)}}{\frac {\left|b^2-c^2\right|}{2a}}$ $\cdot\frac {RD}{RA}\cdot\frac {b+c}{a}=1$ $\Longleftrightarrow$ $\frac {RA}{a}=\frac {RD}{b+c}=\frac {h_a}{2s}$ $\Longleftrightarrow$ $AR=\frac {ah_a}{2s}=\frac {2sr}{2s}$ $\Longleftrightarrow$ $RA=r$ .

Remark. Denote $S\in BC\cap w$ , the diameter $[SN]$ of $w$ and $T\in AN\cap BC$ . Prove easily that $SB=TC=s-b$ ,

i.e. $T\in BC\cap w_a$ , where $w_a$ is $A$-exincircle of $\triangle ABC$ . Also $MS=MT=\frac {|b-c|}{2}$ and $\overline {ANT}\parallel \overline {RIM}$ .



PP3. Let $\triangle ABC$ with incircle $w=C(I,r)$ . Denote midpoint $M$ of $[BC]$ , $D\in BC\cap w$ , $E\in CA\cap w$ , $F\in AB\cap w$ . Prove that $EF\cap DI\cap AM\ne\emptyset$ .

Proof 1. Denote $X\in EF\cap DI$ , $Y\in EF\cap AM$ . Observe that $IE=IF=r$ , $m\left(\widehat{XIE}\right)=C$ and $m\left(\widehat{XIF}\right)=B$ . Using the well-known relation

$\frac {XE}{XF}=\frac {IE}{IF}\cdot\frac {\sin \widehat{XIE}}{\sin\widehat{XIF}}$ obtain that $\boxed{\ \frac {XE}{XF}=\frac cb\ }\ (1)$ . Observe that $MB=MC=\frac a2$ and $AE=AF=s-a$ . Using the well-known relation

$\frac {YE}{YF}=\frac {MC}{MB}\cdot\frac {AE}{AF}\cdot\frac {AB}{AC}$ obtain that $\boxed{\ \frac {YE}{YF}=\frac cb\ }\ (2)$ . From the relations $(1)$ and $(2)$ obtain that $X\equiv Y$ , i.e. $EF\cap DI\cap AM\ne\emptyset$ .

Proof 2. Denote $L\in BC$ for which $AL\perp BC$ , $D\in BC\cap w$ , $Z\in AM\cap DI$ and $Y\in AM\cap EF$ . Observe that $\frac {MA}{MZ}=\frac {ML}{MD}=\frac {\frac {\left|b^2-c^2\right|}{2a}}{\frac {|b-c|}{2}}$ $\Longrightarrow$

$\boxed{\ \frac {MA}{MZ}=\frac {b+c}{a}\ }\ (3)$ . Using the well-known relation $\frac {EC}{EA}\cdot MB+\frac {FB}{FA}\cdot MC=\frac {YM}{YA}\cdot BC$ obtain that $\frac {s-c}{s-a}+\frac {s-b}{s-a}=2\cdot\frac {YM}{YA}$ $\Longleftrightarrow$

$\frac {YA}{YM}=\frac {2(s-a)}{a}$ , i.e. $\boxed{\ \frac {MA}{MY}=\frac {b+c}{a}\ }\ (4)$ . From the relations $(3)$ and $(4)$ obtain that $Z\equiv Y$ , i.e. $EF\cap DI\cap AM\ne\emptyset$ .

Proof 3. Denote $X\in EF\cap DI$ and $M'\in AX\cap BC$ . Observe that $IE=IF=r$ , $m\left(\widehat{XIE}\right)=C$ and $m\left(\widehat{XIF}\right)=B$ .

Using the well-known relation $\frac {XE}{XF}=\frac {IE}{IF}\cdot\frac {\sin \widehat{XIE}}{\sin\widehat{XIF}}$ obtain that $\frac {XE}{XF}=\frac cb$ . Observe that $AE=AF=s-a$ . Using the well-known

relation $\frac cb=\frac {XE}{XF}=\frac {M'C}{M'B}\cdot\frac {AE}{AF}\cdot\frac {AB}{AC}$ obtain that $M'B=M'C$ , i.e. $M'\equiv M$ . In conclusion, $EF\cap DI\cap AM\ne\emptyset$ .



PP4 (nice). Let $\triangle ABC$ with circumcircle $\alpha  =C(O,R)$ and incircle $w=C(I,r)$ . Let $D\in BC\cap w$ , $\{A,S\}=AI\cap \alpha$ , $\{A,A'\}=AO\cap\alpha$ , $L\in SD\cap A'I$ . Prove that $L\in\alpha$ .

Proof. Observe that $\widehat{DIS}\equiv\widehat{IAA'}$ and $\frac {ID}{AI}=\frac {IS}{AA'}$ because the power $p_{\alpha}(I)$ of $I$ w.r.t. $\alpha$ is given by the relation $-p_{\alpha}(I)=IA\cdot IS=$ $2Rr=AA'\cdot ID$ .

Therefore, $\triangle DIS\sim IAA'$ from where obtain $\widehat{DSI}\equiv\widehat{IA'A}$ , i.e. the point $L\in SD\cap A'I$ belongs to the circle $\alpha$ .

Remark. $K\in AI\cap BC$ $\implies$ $ABS\sim AKC\implies$ $\boxed{bc=AK\cdot AS}=AK\cdot (AK+KS)=AK^2+KB\cdot KC\implies$ $\boxed{KA^2=bc-KB\cdot KC}$ .

Prove easily that $AK\cdot AS=\boxed{bc=AI\cdot AI_a}$ and show analogously that $K'\in BC$ for which $AK'\perp AK\implies$ $\boxed{K'A^2=K'B\cdot K'C-bc}$ .
This post has been edited 47 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:43 PM

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