Track / Stage 6 / 205 of 400 #1205 of 1964

Problem 1205

National olympiad, first round

Geometry Difficulty 6.3 Prove it

13. As shown in Figure $1, P A$ and $P B C$ are the tangent and secant of $\odot O$ , respectively, with $A$ being the point of tangency, and $M$ being the midpoint of the tangent $P A$ . Chord $A D$ intersects $B C$ at point $E$ , and point $F$ on the extension of chord $A B$ satisfies $\angle F B D=\angle F E D$ . Prove that the necessary and sufficient condition for $P, F, D$ to be collinear is that $M, B, D$ are collinear.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

13. From $P A$ being the tangent of $\odot O$ , we know
$\begin{array}{l} \angle P A D + \angle A B D = 180^{\circ}. \\ \text { Also, } \angle F B D + \angle A B D = 180^{\circ}, \text { then } \\ \angle P A D = \angle F B D = \angle F E D \\ \Rightarrow E F \parallel A P. \end{array}$
(1) If $M, B, D$ are collinear, let the intersection of line $A B$ and $D P$ be point $F_{1}$ .
By Ceva's Theorem, we have $\frac{A M}{M P} \cdot \frac{P F_{1}}{F_{1} D} \cdot \frac{D E}{E A} = 1$ .
Noting that $A M = M P$ ,
then $\frac{P F_{1}}{F_{1} D} = \frac{A E}{E D}$ , and $E F_{1} \parallel A P$ .
Since points $F$ and $F_{1}$ are both on line $A B$ , point $F$ coincides with $F_{1}$ .
Therefore, $P, F, D$ are collinear.
(2) If $P, F, D$ are collinear, let the intersection of line $D B$ and $A P$ be point $M_{1}$ .
By Ceva's Theorem, we have $\frac{A M_{1}}{M_{1} P} \cdot \frac{P F}{F D} \cdot \frac{D E}{E A} = 1$ .
Since $E F \parallel A P$ , $\frac{P F}{F D} = \frac{A E}{E D}$ , then
$A M_{1} = M_{1} P,$

which means $M_{1}$ is the midpoint of $P A$ . Hence, point $M_{1}$ coincides with $M$ .
Therefore, $M, B, D$ are collinear.

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