Banguizb

Answered

2022-06-30

Let $\mathrm{\u25b3}ABC$ and E, D on $[AB]$ and $[AC]$ s.t. BEDC is inscribable. Let $P\in [BD],Q\in [CE]$ , s.t. AEPC and ADQB are also inscribable. Show that $AP=AQ$ .

Answer & Explanation

zlepljalz2

Expert

2022-07-01Added 22 answers

Step 1

$\mathrm{\angle}ABD=\mathrm{\angle}EBD=\mathrm{\angle}ECD=\mathrm{\angle}ACE\Rightarrow $

$\mathrm{\u25b3}ABD\sim \mathrm{\u25b3}ACE\Rightarrow $

$\frac{AD}{AB}=\frac{AE}{AC}\Rightarrow $

$AD\cdot AC=AB\cdot AE$

$\mathrm{\angle}EPA=\mathrm{\angle}ECA=\mathrm{\angle}ECD=\mathrm{\angle}EBD=\mathrm{\angle}PBA\Rightarrow $

$\mathrm{\u25b3}EPA\sim \mathrm{\u25b3}PBA\Rightarrow $

$\frac{AE}{AP}=\frac{AP}{AB}\Rightarrow $

$A{P}^{2}=AB\cdot AE$

In the same way: $A{Q}^{2}=AC\cdot AD$

Then $A{P}^{2}=AB\cdot AE=AC\cdot AD=A{Q}^{2}$ , $AP=AQ$

$\mathrm{\angle}ABD=\mathrm{\angle}EBD=\mathrm{\angle}ECD=\mathrm{\angle}ACE\Rightarrow $

$\mathrm{\u25b3}ABD\sim \mathrm{\u25b3}ACE\Rightarrow $

$\frac{AD}{AB}=\frac{AE}{AC}\Rightarrow $

$AD\cdot AC=AB\cdot AE$

$\mathrm{\angle}EPA=\mathrm{\angle}ECA=\mathrm{\angle}ECD=\mathrm{\angle}EBD=\mathrm{\angle}PBA\Rightarrow $

$\mathrm{\u25b3}EPA\sim \mathrm{\u25b3}PBA\Rightarrow $

$\frac{AE}{AP}=\frac{AP}{AB}\Rightarrow $

$A{P}^{2}=AB\cdot AE$

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

2022-07-02Added 2 answers

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.

$\mathrm{\angle}AQR\ne \mathrm{\angle}ADP$ 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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