Question

Prove that the centre of the circum circle of a right angled triangle is the midpoint of its hypotenuse.

Answer 1

Given: A right-angled ΔACB with O as the midpoint of hypotenuse AB

Construction: Draw OD ⊥ AC. Join OC.

∠ACB = ∠ADO = 90°

So, by converse of corresponding angle axiom, DO || CB.

We know that in a triangle, a line through the midpoint of a side and parallel to another side bisects the third side.

⇒ AD = DC

Now, in ΔAOD and ΔCOD:

DO = DO (Common side)

AD = DC (Proved above)

∠ADO = ∠CDO = 90°

∴ ΔAOD ≅ ΔCOD (By side angle side criterion of congruence)

⇒ AO = OC (Corresponding parts of congruent triangles are congruent)

O is the midpoint of AB.

∴ OB = AO

⇒ OB = AO = OC

Thus, O is the centre of the circle passing through points A, B and C.

Hence, the centre of the circumcircle of a right-angled triangle is the midpoint of its hypotenuse.