Austrian Mathematical Olympiad

Question

Let ABC be an isosceles triangle with AC = BC and circumcircle k. The point D lies on the shorter arc of k over the chord BC and is different from B and C. Let E denote the intersection of CD and AB. Prove that the line through B and C is a tangent of the circumcircle of the triangle BDE. FIGURE_0

Accepted Answer

We denote the center of the circumcircle of the triangle

B D E

by

M

and

∠ B A C = ∠ C B A

by

α

. Since the quadrilateral

A B D C

is cyclic, we obtain

∠ B D E = α

. By the inscribed angle theorem,

∠ B M E = 2 α

and thus

∠ E B M = ∠ M E B = 9 0^{\circ} - α

. Therefore,

18 0^{\circ} = ∠ C B A + ∠ M B C + ∠ E B M = α + ∠ M B C + 9 0^{\circ} - α = ∠ M B C + 9 0^{\circ}

holds and we get

∠ M B C = 9 0^{\circ},

completing the proof.

Figure 1: Problem 6