Track / Stage 5 / 308 of 400 #908 of 2000

Problem 908

AIME late

Geometry Difficulty 5.8 Prove it

II. (Total 25 points) As shown in Figure 4, $P$ is the midpoint of line segment $AB$ , $\angle BPC = \angle BPD$ , and $PB$ is the mean proportional between $PC$ and $PD$ . If points $A$ , $B$ , and $C$ are all on circle $\odot K$ , then is point $D$ inside $\odot K$ , on $\odot K$ , or outside $\odot K$ ? Prove your conclusion.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

II. Conclusion: Point $D$ is on
$\odot K$ , i.e., points $A$ , $B$ , $C$ ,
$D$ are concyclic. Auxiliary lines see Figure 8.
It is easy to know that $\triangle P A E \cong$
$\triangle P B C$ , then $A B C E$ is
an isosceles trapezoid, which must have a circumcircle, i.e., points $A$ , $B$ , $C$ , $E$ are concyclic.
Also, $\because P E \cdot P D=P C \cdot P D=P B^{2}=P A \cdot P B$ ,
$\therefore A$ , $D$ , $B$ , $E$ are concyclic.
In the two sets of four concyclic points $A$ , $B$ , $C$ , $E$ and $A$ , $D$ , $B$ , $E$ , there are three common points $A$ , $B$ , $E$ , which can determine a circle, hence points $A$ , $B$ , $C$ , $D$ , $E$ must lie on the same circle, i.e., points $A$ , $B$ , $C$ , $D$ are concyclic.

Alternative proof hint: Using $\triangle P A C \backsim \triangle P D A$ , $\triangle P B C \backsim$ $\triangle P D B$ , it is not difficult to deduce that $\angle D A C$ and $\angle D B C$ are supplementary.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.