Question

In a right triangle, prove that the square on the hypotenuse is equal to sum of the squares on the other two sides

Answer 1

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,

$△ADB\sim △ABC.$ (by AA similarity)

Therefore, $AD/AB=AB/AC$

(In similar Triangles corresponding sides are proportional)

$AB²=AD\times AC$……..$\left(1\right)$

Also, $△BDC\sim △ABC$

Therefore, $CD/BC=BC/AC$

(in similar Triangles corresponding sides are proportional)

Or, $BC²=CD\times AC$……..$\left(2\right)$

Adding the equations $\left(1\right)$ and $\left(2\right)$ we get,

$AB²+BC²=AD\times AC+CD\times AC$

$AB²+BC²=AC(AD+CD)$

( From the figure $AD+CD=AC$)

$AB²+BC²=AC.AC$

Therefore, $AC²=AB²+BC²$

This theorem is known as Pythagoras theorem...