 Banguizb

2022-06-30

Let $\mathrm{△}ABC$ and E, D on $\left[AB\right]$ and $\left[AC\right]$ s.t. BEDC is inscribable. Let $P\in \left[BD\right],Q\in \left[CE\right]$ , s.t. AEPC and ADQB are also inscribable. Show that $AP=AQ$ . zlepljalz2

Expert

Step 1
$\mathrm{\angle }ABD=\mathrm{\angle }EBD=\mathrm{\angle }ECD=\mathrm{\angle }ACE⇒$
$\mathrm{△}ABD\sim \mathrm{△}ACE⇒$
$\frac{AD}{AB}=\frac{AE}{AC}⇒$
$AD\cdot AC=AB\cdot AE$
$\mathrm{\angle }EPA=\mathrm{\angle }ECA=\mathrm{\angle }ECD=\mathrm{\angle }EBD=\mathrm{\angle }PBA⇒$
$\mathrm{△}EPA\sim \mathrm{△}PBA⇒$
$\frac{AE}{AP}=\frac{AP}{AB}⇒$
$A{P}^{2}=AB\cdot AE$
In the same way: $A{Q}^{2}=AC\cdot AD$
Then $A{P}^{2}=AB\cdot AE=AC\cdot AD=A{Q}^{2}$ , $AP=AQ$ rmd1228887e

Expert

Step 1
$\mathrm{\angle }AQR\ne \mathrm{\angle }ADP$ as mentioned in comment.I made two drawings, one isosceles. These points are noticeable:
The measure of the sides and angles of isosceles triangle APQ does not vary with the type of triangle.
It can be assumed as a kind of affine transformation. Because base of triangle is fixed and the location of vertex A alters. It is like moving a triangle APQ relative to the base of ABC to construct three circles with certain measures of diameter.

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