Question

The median AM of triangle ∆ABC is half the length of the side towards which it is drawn, BC . Prove that triangle ∆ABC is a right triangle.

Answer 1

The problem statement tells us that

• AM ≅ MB, ∴ ΔAMB is isosceles
• AM ≅ MC, ∴ ΔAMC is isosceles

Base angles of an isosceles triangle are equal, so ∠MAB ≅ ∠MBA. ∠AMC is the exterior angle opposite those two, so it is equal to their sum, 2∠MAB.

The base angles in isosceles ΔAMC are equal to half the difference between the apex angle, ∠AMC, and 180°. That is,

... ∠MAC = (1/2)(180° -∠AMC) = (1/2)(180° -2∠MAB) = 90° -∠MAB

The angle at A of ΔABC is the sum of the two angles created by the median AM. That is ...

∠A = ∠MAC + ∠MAB

∠A = (90° -∠MAB) +∠MAB

∠A = 90°

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Maybe a shorter way to get there is to realize that ...

... MA ≅ MB ≅ MC

so M is the center of a circle with BC as a diameter and A a point on the circle. Angle BAC is inscribed in the semicircle and subtends an arc of 180°, so angle A is 90°.