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Estonian Math Competitions

Geometry Difficulty 4.8 AIME Prove it Estonia

A right triangle $ABC$ has the right angle at vertex $A$ . Circle $c$ passes through vertices $A$ and $B$ of the triangle $ABC$ and intersects the sides $AC$ and $BC$ correspondingly at points $D$ and $E$ . The line segment $CD$ has the same length as the diameter of the circle $c$ . Prove that the triangle $ABE$ is isosceles.

Solutions — 2

Solution 1

Since $\angle BAD = 90^\circ$ (Fig. 1), $BD$ is the diameter of circle $c$ and therefore $CD = BD$ . Since $BD$ is diameter, also $\angle BED = 90^\circ$ , so $DE$ is an altitude of the isosceles triangle $BDC$ , bisecting its base $BC$ . Hence $E$ is the midpoint of the hypotenuse $BC$ of the triangle $ABC$ . Since the midpoint of the hypotenuse is the circumcentre of a right triangle, it follows $EA = EB$ . This means that $ABE$ is an isosceles triangle.

Fig. 1

Solution 2

As in the previous solution, we show that $CD = BD$ . Hence $\angle ECD = \angle EBD$ . From the equality of the inscribed angles subtending the arc ED it also follows $\angle EBD = \angle EAD$ . From the triangle ABC we get $\angle ABC = 90^\circ - \angle BCA$ , or $\angle EBA = 90^\circ - \angle ECD$ . On the other hand, $\angle EAB = \angle DAB - \angle EAD = 90^\circ - \angle EBD = 90^\circ - \angle ECD$ . Consequently $\angle EBA = \angle EAB$ . So the triangle ABC is isosceles.

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