In $△ A B C, m ∠ B = 90^{o}$ . Prove that ${A C}^{2} = {A B}^{2} + {B C}^{2}$ .

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Solution

Given: In $△ A B C, m ∠ B = 90^{o}$

To prove: ${A C}^{2} = {A B}^{2} + {B C}^{2}$

Construction: Draw $B D ⊥ A C$

Proof:
We know that, if a perpendicular is drawn from the vertex of the right angle of a right angled triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

$△ A D B \sim △ A B C$

If two triangles are similar, then their corresponding sides are proportional.

So, $\frac{A D}{A B} = \frac{A B}{A C}$

or $A D . A C = {A B}^{2} . . . . (1)$

Also, $△ B D C \sim △ A B C$

Similarly, $\frac{C D}{B C} = \frac{B C}{A C} C D . A C = {B C}^{2} . . . (2)$

Adding $(1)$ and $(2)$

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

$A C (A D + C D) = {A B}^{2} + {B C}^{2}$

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

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

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