Question

In triangle ABC, angle C = 90°, and AD, BE are medians through A and B respectively. Prove that
4AD²= 4AC² + BC²​

Answer 1

Figure is in the attachment...

Given

In triangle ABC, angle C = 90° and AD, BE are medians through A and B respectively.

We have to prove ; 4AD² = 4AC² + BC²

Proof

Here AD is median of BC, so CD = ½ BC.. (1)

In ∆ACD ; applying Pythagoras theorem ;

→ AD² = AC² + CD²........ (2)

Now putting values of CD in (2) :

→ AD² = AC² + (BC/2)² [CD = BC/2]

→ AD² = AC² + (BC²/4)

→ AD² = (4AC² + BC²)/4

After cross multiplication :

→ 4AD² = 4AC² + BC²

Therefore, L.H.S. = R.H.S [Proved]