Question

If $△ABC$ is right-angled at$B$. then prove that ${AB}^2+{BC}^2=A{C}^2$

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, then triangle on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

$△ADB\sim △ABC$

$\therefore \frac{AD}{AB}=\frac{AB}{AC}$ (In similar triangles corresponding sides are equal)

$A{B}^2=AD\times AC.........\left(1\right)$

Also $△BDC\sim △ABC$

$\therefore \frac{CD}{BC}=\frac{BC}{AC}$ (In similar triangles corresponding sides are equal)

$B{C}^2=CD\times AC.........\left(2\right)$

Adding (1) and (2) we get

$A{B}^2+B{C}^2=AD\times AC+CD\times AC$

$A{B}^2+B{C}^2=AC(AD+A=CD)$

From the figure $AD+CD=AC$

$A{B}^2+B{C}^2=AC.AC$

Therefore, $A{B}^2+B{C}^2=A{C}^2$