Two circles Γ and Σ in the plane intersect at two distinct points A and B and the centre of Σ lies on Γ . Let points C and D be on Γ and Σ, respectively, such that C, B and D are collinear. Let point E on Σ be such that DE is parallel to AC. Show that AE = AB.

## Comments

Sort by:

TopNewestLet \(O\) be the centre of Γ, and let \(P\) be the centre of Σ. Let the point \(F\) be on Γ, such that \(F,O,P\) are collinear. Let \(\angle ACB = x\).

From the picture, we can see that there are two similar possibilities for how the diagram turns out. By properties of parallel lines, \(\angle BDE\) either equals \(x\) or \(180^{\circ} - x\). In both cases, since \(ABDE\) is a cyclic quadrilateral, this implies that \(\angle BAE = x = \angle ACB\).

[1]Since \(FCBA\) is a cyclic quadrilateral, we have \(\angle ACB = \angle AFB\). Combining this with

[1]gives us \(\angle BAE = \angle BFA\).[2]Since \(F,O,P\) are collinear, then \(FP\) is a diameter of Γ. Therefore \(\angle FAP = \angle FBP = 90^{\circ}\).

[3]Since \(PA\) and \(PB\) are radii of Σ, we have \(PA = PB\). Therefore the triangles \(\triangle FPA\) and \(\triangle FPB\) have two pairs of sides equal (\(PA = PB\) and \(FP = FP\)) and one pair of right angles. Hence \(\triangle FPA\) and \(\triangle FPB\) are congruent. Hence \(FA = FB\).

[4]Since \(P\) is the center of Σ and \(E\) is a point on Σ on the same side of \(AB\) as \(P\), therefore \(\angle BPA = 2\angle BEA\). Now since \(\triangle BPA\) is an isosceles triangle, and since the angles in a triangle sum to \(180^{\circ}\), we have \(\angle BAP = \dfrac{180^{\circ} - \angle BPA}{2} = 90^{\circ} - \angle BEA\). But thanks to

[3], we know that \(\angle FAB = 90^{\circ} - \angle BAP\). Hence, \(\angle FAB = \angle BEA\).[5]Therefore, by

[2]and[5], the triangles \(\triangle BEA\) and \(\triangle FAB\) have two pairs of equal angles, which means they have three pairs of equal angles. Therefore \(\triangle BEA\) and \(\triangle FAB\) are similar triangles.Therefore, by

[4], since \(FA = FB\), then \(AE = AB\) also.QED– Ariel Gershon · 1 month agoLog in to reply

thats a very good solution you are the best – Shashwat Gokhe · 1 month ago

Log in to reply

– Ariel Gershon · 1 month ago

Thanks! This was a very interesting problem.Log in to reply