Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National Olympiad Prove it Philippines

Problem:
Let $P$ be a point outside a circle, and let the two tangent lines through $P$ touch the circle at $A$ and $B$ . Let $C$ be a point on the minor arc $A B$ , and let $\overrightarrow{P C}$ intersect the circle again at another point $D$ . Let $L$ be the line that passes through $B$ and is parallel to $\overline{P A}$ , and let $L$ intersect $\overrightarrow{A C}$ and $\overrightarrow{A D}$ at points $E$ and $F$ , respectively. Prove that $B$ is the midpoint of $\overline{E F}$ .

Solution

Solution:
Refer to Figure 9. We only need to see four pairs of similar triangles.
$\begin{aligned} \triangle A E B \sim \triangle A B C & \Longrightarrow \frac{B E}{A B}=\frac{B C}{A C} \\ \triangle A F B \sim \triangle A B D & \Longrightarrow \frac{B F}{A B}=\frac{B D}{A D} \\ \triangle P B C \sim \triangle P D B & \Longrightarrow \frac{B C}{B P}=\frac{B D}{D P} \quad \Longrightarrow \quad \frac{B C}{B D}=\frac{B P}{D P} \\ \triangle P A C \sim \triangle P D A & \Longrightarrow \frac{A C}{A P}=\frac{A D}{D P} \quad \Longrightarrow \quad \frac{A C}{A D}=\frac{A P}{D P} \end{aligned}$

Figure 9: Problem 3.
Since $A P=B P$ , from $(3 \star)$ and $(4 \star)$ , we get
$\frac{B C}{B D}=\frac{A C}{A D} \quad \text{ or } \quad \frac{B C}{A C}=\frac{B D}{A D}$
Using this last proportion and applying transitivity property to ( $1 \star$ ) and $(2 \star)$ yield
$\frac{B E}{A B}=\frac{B F}{A B} \quad \text{ or } \quad B E=B F$

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