Question

ABC is an isosceles triangle in which AB=AC and BC square=2AB square then prove that ABC is a right triangle

Answer 1

You can prove this by using the cosine rule.

In any triangle, $a^2 = b^2 + c^2 - 2bcCosA$

Where a, b, c are sides of a triangle and A is the angle opposite to side a.

So taking in triangle ABC, the side opposite to angle A will be BC.

BC=a. Similarly, AC = b and AB = c.

We know that AB=AC

So b=c

By cosine rule,

$a^2= b^2 +c^2 -2bcCosA\\CosA = (2b^2-a^2)/ 2bc$

It is given that $a^2 = 2b^2$

$CosA= 0\\A=90$

Therefore, the triangle is right angled

Answer 2

data:BC square =2 AB square
AB= AC
to prove :abc is a right triangle
proof : in ABC,
BC square =2 AB square. (data)
BC square = AB square + AB square
BC square =AB square + AC square (ab =ac)
therefore, BAC is a right triangle