Question

in triangle ABC, angle B= 90 and M is a point on BC . Prove that AM^2+BC^2=AC^2+BM^2

Answer 1

Step-by-step explanation:

given :

in triangle ABC, angle B= 90 and M is a point on BC . Prove that AM^2+BC^2=AC^2+BM^2

to find :

Prove that AM^2+BC^2=AC^2+BM^2

solution :

• 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 AABM and applying Pythagoras theorem we get,

• AM² = = AB + BM²

• AB²= AM² - BM²

• .(i)

• Now, we consider the AABC and applying Pythagoras theorem we get,

• AC² = AB2 + BC²

• AB² = AC²-BC²

• From (i) and (ii) we get, AM² - BM² = AC² - BC² AM² + BC2 = AC² + BM²