Prove that in a right angled triangle square of hypotenuse is egual to the sum of sguares of its other two sides
Answers
AC²=AB²+BC²
Given: A right angled ∆ABC, right angled at B
To Prove: AC²=AB²+BC²
Construction: Draw perpendicular BD onto the side AC .
Proof:
We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
We have
△ADB∼△ABC. (by AA similarity)
Therefore, AD/ AB=AB/AC
(In similar Triangles corresponding sides are proportional)
AB²=AD×AC……..(1)
Also, △BDC∼△ABC
Therefore, CD/BC=BC/AC
(in similar Triangles corresponding sides are proportional)
Or, BC²=CD×AC……..(2)
Adding the equations (1) and (2) we get,
AB²+BC²=AD×AC+CD×AC
AB²+BC²=AC(AD+CD)
( From the figure AD + CD = AC)
AB²+BC²=AC . AC
Therefore, AC²=AB²+BC²
Hope this helps you!!!