114. A property of tangential triangle for ABC.

by Virgil Nicula, Sep 12, 2010, 3:16 AM

Quote:
Let $ABC$ be a triangle. The tangents to its circumcircle at $A$, $B$, $C$ form a triangle $PQR$ with $C \in PQ$

and $B \in PR$ . Let $D$ be the foot of the altitude from $C$ in $ABC$. Prove that $CD$ bisects $\angle QDP$ .

Proof. I note $T\in AB\cap CC,\ S\in AB\cap CR$. Because $CD\perp AB$, then the ray $[CD$ bisects the angle $\angle PDQ$ iff the division $(TQCP)$ is harmonically.

It is well-known that the ray $[CS$ is the simedian from the vertex $C$ of the triangle $ABC$ and $\frac{TA}{TB}=\frac{SA}{SB}=\left( \frac ba \right)^2$. Thus, $R(T,A,S,B)$ is harmonical pencil $\Longrightarrow$

$PC\cap R(T,A,S,B)$ is harmonical division $\Longrightarrow (T,Q,C,P)$ is harmonical division. Remark that if denote $m(\widehat{CDQ})=m(\widehat{CDP})=\phi$ , then prove easily that

$\boxed{\tan\phi =\frac {1+\tan A\tan B}{\tan A+\tan B}}$ . Indeed, appling the Sinus' theorem in $\triangle CDP$ obtain that $\frac {DC}{\cos (A-B+\phi )}=\frac {CP}{\sin\phi}$ $\iff$ $\frac {a\sin B}{\cos (A-B+\phi )}=\frac {\frac {a}{2\cos A}}{\sin\phi}$ $\iff$

$2\cos A\sin B\sin\phi =\cos (A-B+\phi )$ $\iff$ $2\cos A\sin B\tan\phi =\cos (A-B)-\sin (A-B)\tan\phi$ $\iff$ $\tan\phi =\frac {\cos (A-B)}{\sin (A-B) +2\cos A\sin B}$

$\iff$ $\tan\phi =\frac {\cos (A-B)}{\sin (A+B)}$ $\iff$ $\tan\phi =\frac {1+\tan A\tan B}{\tan A+\tan B}$ . I used in the proof of the proposed problem a well-known property of the harmonical division :
Quote:
Let $A,B,C,D\in d$ be four points which belong to the same line $d$ (in the mentioned order) and the point $P\not\in d\ .$

Then if two from the following affirmations are truly, results that and the other affirmation is truly:

$\blacksquare \ 1.\ (A,B,C,D)$ - harmonical division $\left(\ \frac{\overline {BA}}{\overline {BC}}=-\frac{\overline {DA}}{\overline {DC}}\ \right)\ ;$

$\blacksquare \ 2.\ PB\perp PD\ ;$

$\blacksquare \ 3.$ The ray $[PB$ bisects the angle $\angle APC\ .$
Quote:
Particular case. Let $ABC$ be a $C$-right triangle with circumcircle $w$ . Denote $D\in (AB)$ so that $CD\perp AB$ , $P\in BB\cap CC$ and $Q\in AA\cap CC$ .

Prove that $m(\widehat{CDP})=m(\widehat{CDQ})=\phi$ and $\tan\phi =\frac {2h_c}{c}$ . I used the notation $XX$ for the tangent line to the circumcircle $w$ in the point $X\in w$ .
Quote:
A similar problem. Let $ABC$ be a triangle, the its circumcircle $w=C(O,R)$, the reflection $A'$ of the point $A$ w.r.t. the center $O$,

the point $T\in BC\cap A'A'$ and the proiection $P$ of the point $A'$ on the line $OT$. Prove that the ray $[PA'$ bisects the angle $\angle BPC$.

Indication. $I\in BC\cap PA'\Longrightarrow (B,I,C,T)$ is a harmonical division.
This post has been edited 32 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:49 AM

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