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Geometry Difficulty 5.1 AIME, harder Prove it Estonia

Two circles cc and c′c' with centers OO and O′O' lie completely outside each other. Points AA, BB, and CC lie on the circle cc and points A′A', B′B', and C′C' lie on the circle c′c' so that segment AB∥A′B′AB \parallel A'B', BC∥B′C′BC \parallel B'C', and ∠ABC=∠A′B′C′\angle ABC = \angle A'B'C'. The lines AA′AA', BB′BB', and CC′CC' are all different and intersect in one point PP, which does not coincide with any of the vertices of the triangles ABCABC or A′B′C′A'B'C'. Prove that ∠AOB=∠A′O′B′\angle AOB = \angle A'O'B'.

Figure 1
Fig. 1

Solution

The triangles ABPABP and A′B′PA'B'P are similar, because their corresponding sides are parallel (Fig. 1). Hence ∣AB∣∣A′B′∣=∣BP∣∣B′P∣\frac{|AB|}{|A'B'|} = \frac{|BP|}{|B'P|}. Likewise the triangles BCPBCP and B′C′PB'C'P are similar, hence ∣BC∣∣B′C′∣=∣BP∣∣B′P∣\frac{|BC|}{|B'C'|} = \frac{|BP|}{|B'P|}. Thus ∣AB∣∣A′B′∣=∣BC∣∣B′C′∣\frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|}, and since ∠ABC=∠A′B′C′\angle ABC = \angle A'B'C', the triangles ABCABC and A′B′C′A'B'C' are also similar. From the equality of the angles ACBACB and A′C′B′A'C'B' the equality of the central angles AOBAOB and A′O′B′A'O'B' now follows.

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