Question

ABC is a triangle, right-angled at B. M is a point on BC.

Prove that: AM² + BC² = AC² + BM²

Answer 1

The pictorial form of the given problem is as follows:

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔABM and applying Pythagoras theorem we get,

AM² = AB² + BM² 

AB² = AM² - BM²               ...(i)

Now, we consider the ΔABC and applying Pythagoras theorem we get,

AC² = AB² + BC² 

AB² = AC² - BC²                ...(ii)

From (i) and (ii) we get,

AM² - BM² = AC² - BC² 

AM² + BC² = AC² + BM²  

Hence, Proved.