Question

ABC is an isosceles triangle right angled at C, prove that AB^2 =2AC^2?

Answer 1

As $ABC$ is a right angle triangle than we can say $AB$ is the hypotenuse, $AC$ is the base and similarly altitude be $BC$.

So, the Pythagorean theorem : $(Hyp)^2 = (Base)^2 + (Altitude )^2$

$=>AB^2=AC^2+BC^2$..................(1)
As the triangle is isosceles 2 sides will be same.
Thus, $AC=BC$

Putting this to (1) we get

$AB^2=AC^2+AC^2$
so, $AB^2=2AC^2(proved)$

Answer 2

ur answer is attached....

hope it's helpful for u....