Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Given:- A right angled triangle ABC, right angle at B
To prove:- AC²=AB²+BC²
Proof:- Draw a perpendicular BD from B to AC
In ΔABC and ΔABD
∠ADB = ∠ABC = 90°
∠DAB = ∠BAC. (Common angle)
∴ΔABC ≅ ΔABD.. (Using AA similarity criteria)
Now, AD/AB = AB/AC. (Corresponding sides are proportional)
⇒AB²=AD × AC. (i)
Similarly, ΔABC ≅ ΔBDC
∴BC²=CD × AC. (ii)
Adding equations (i) and (ii), we get
AB²+BC²=AD × AC + CD × AC
⇒AB²+BC²=AC(AD+CD)
⇒AB²+BC²=AC × AC
⇒AB²+BC²=AC² [hence proved]