Track / Stage 7 / 159 of 300 #1559 of 2000

Problem 1559

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.3 Prove it

Let $ABC$ be a triangle. Let $D$ and $E$ be respectively points on the segments $AB$ and $AC$ , and such that $DE||BC$ . Let $M$ be the midpoint of $BC$ . Let $P$ be a point such that $DB=DP$ , $EC=EP$ and such that the open segments (segments excluding the endpoints) $AP$ and $BC$ intersect. Suppose $\angle BPD=\angle CME$ . Show that $\angle CPE=\angle BMD$

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

1. Define the circles and intersection point:
- Let $\Gamma_B \equiv \odot(D, DB)$ and $\Gamma_C \equiv \odot(E, EC)$ .
- Let $Q$ be the second intersection of $\Gamma_B$ and $\Gamma_C$ .

2. **Reflect $C$ in $E$ :**
- Let $X$ be the reflection of $C$ in $E$ .

3. Establish the perpendicularity:
- Since $DE \parallel BC$ , and $DE \perp PQ$ , it follows that $BC \perp PQ$ .

4. **Angle relationships involving $D$ , $P$ , and $B$ :**
- Since $D$ is the circumcenter of $\triangle PQB$ , we have $\measuredangle DPB = 90^\circ - \measuredangle BQP = \measuredangle CBQ$ .

5. **Angle relationships involving $C$ , $M$ , and $E$ :**
- Since $ME$ is the C-midline of $\triangle BXC$ , we have $\measuredangle CME = \measuredangle CBX$ .

6. Equating the angles:
- Therefore, $\measuredangle DPB = \measuredangle CME \iff \measuredangle CBQ = \measuredangle CBX \iff B \in QX$ .

7. Using the diameter property:
- Since $\overline{CX}$ is a diameter of $\Gamma_C$ , we have $\measuredangle CQX = 90^\circ$ .

8. Conclusion:
- Hence, $\measuredangle DPB = \measuredangle CME \iff \measuredangle BQC = 90^\circ$ .

9. **Analogous argument for $\measuredangle EPC$ and $\measuredangle BMD$ :**
- Similarly, $\measuredangle EPC = \measuredangle BMD \iff \measuredangle BQC = 90^\circ$ .

10. Final conclusion:
- Since both conditions lead to $\measuredangle BQC = 90^\circ$ , we have shown that $\measuredangle CPE = \measuredangle BMD$ .

$\blacksquare$

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