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

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Solution

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
$△ A D B \sim △ A B C$
$∴ \frac{A D}{A B} = \frac{A B}{A C}$ (In similar triangles corresponding sides are equal)
$A B^{2} = A D \times A C . . . . . . . . . (1)$
Also $△ B D C \sim △ A B C$
$∴ \frac{C D}{B C} = \frac{B C}{A C}$ (In similar triangles corresponding sides are equal)
$B C^{2} = C D \times A C . . . . . . . . . (2)$
Adding (1) and (2) we get
$A B^{2} + B C^{2} = A D \times A C + C D \times A C$
$A B^{2} + B C^{2} = A C (A D + A = C D)$
From the figure $A D + C D = A C$
$A B^{2} + B C^{2} = A C . A C$
Therefore, $A B^{2} + B C^{2} = A C^{2}$

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