Question

In the figure below, ABC is a right angled triangle and M is the midpoint of AB.

(a) Prove that .

(b) Prove that MC = MA = MB.

Answer 1

In ΔABC, M is the midpoint of AB.

⇒ BM = MA = …(1)

Also, ∠MNB = ∠ACB = 90°

By converse of corresponding angle axiom, MN || AC

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

⇒ BN = NC = …(2)

Applying Pythagoras theorem in ΔMNB:

MB² = MN² + BN²

⇒ MN² = MB² − BN²

Applying Pythagoras theorem in ΔACB:

AB² = AC² + BC²

⇒ AC² = AB² − BC²

Putting the value of AB² − BC² in equation (3):

Now, in ΔMNC and ΔMNB:

BN = NC (From equation (2))

∠MNB = ∠MNC = 90° (Given)

MN = MN (Common side)

∴ ΔMNC ≅ ΔMNB (By side angle side criterion of congruence)

⇒ MC = MB (Corresponding parts of congruent triangles are congruent)

Thus, we have MC = MA = MB.