Question

in quadilateral ABCD CA=CD,angleB=90°and AD^2=AB^2+BC^2 +CA^2.
Prove that angleACD=90°

Answer 1

Here,

Given that : - AC = CD

AD² = AB² + BC² + CA²

To prove : - Angle ACD = 90°

Proof : -

Now,

In ∆ABC, By Pythagoras theorem,

AC² = AB² + AC² ......(i)

Again,

AD² = AB² + BC² + CA² [Given]

=> AD² = (CA)² + CA² [From (i)]

=> AD² = (CD)² + CA² [CD = CA , given ]

Thus ,

In ∆ACD, AD² = CD² + CA²

So, by converse of Pythagoras theorem,

∆ACD is a right angled triangle ,where AD is the hypotenuse.

Then , angle opposite to hypotenuse is a right angle.

Angle opposite to AD = Angle ACD

Therefore, Angle ACD = 90°

Hence, proved.