In a convex quadrilateral $ABCD$, $P$ and $Q$ are points on sides $BC$ and $DC$ such that $B\hat{A}P = D\hat{A}Q$. If the line that passes through the orthocenters of $\triangle ABP$ and $\triangle ADQ$ is perpendicular to $AC$, prove that the area of these triangles are equals.
Problem
Source:
Tags: geometry
Thelink_20
22.06.2024 03:57
Let $H_1$ and $H_2$ be the orthocenters, $E$ and $F$ be the feet of the altitudes from $A$ and $T=H_1H_2\cap AC$. $[ABP]=[ADQ]\iff\frac{AE}{AF}=\frac{DQ}{BP} (*)$ Because $AC\perp H_1H_2$, we have $CTH_1E$ and $CTH_2F$ cyclic and $A$ lies on their radical axis, thus: $AH_1\cdot AE=AH_2\cdot AF\iff\frac{AE}{AF}=\frac{AH_2}{AH_1}\rightarrow (*)\iff\frac{AH_1}{AH_2}=\frac{BP}{DQ}$
Let $D, E, F$ be the feet of the heights on triangle $ABC$, and $H$ its orthocenter.
$\frac{AE}{AH}=\sin\angle AHE=\sin\angle ACB\Rightarrow AH=\frac{AB\cdot\cos\angle BAC}{\sin\angle ACB}$
$\rightarrow$Then: $AH_1=\frac{AB\cdot\cos\angle BAP}{\sin\angle APB}, AH_2=\frac{AD\cdot\cos\angle DAQ}{\sin\angle AQD}=\frac{AD\cdot\cos\angle BAP}{\sin\angle AQD}$ $\Rightarrow\frac{AH_1}{AH_2}=\frac{AB}{AD}\cdot\frac{\sin\angle AQD}{\sin\angle APB}$ On the other hand, by the Law of Sines: $BP=AB\cdot\frac{\sin\angle BAP}{\sin\angle APB}, DQ=AD\cdot\frac{\sin\angle DAQ}{\sin\angle AQD}=AD\cdot\frac{\sin\angle BAP}{\sin\angle AQD}$ $\Rightarrow\frac{BP}{DQ}=\frac{AB}{AD}\frac{\sin\angle AQD}{\sin\angle APB}=\frac{AH_1}{AH_2}$, as desired!
Attachments:
