Question

Prove that in a right angled triangle square of the hypotenuse is equal to sum of the squares

of other two sides.

Answer 1

Answer:

Step-by-step explanation:

It is your answers

Answer 2

Answer:

Given: Δ ABC is right angled Triangle, m∠B = 90°

To prove: AC² = AB² + CB²

Construction: Draw BD ⊥ AC

Figure is attached.

In ΔABD and Δ ABC

∠A = ∠A  ( common angle )

∠ADB = ∠ABC ( right angle )

Thus, ΔABD and Δ ABC are similar by AA similarity rule.

So,

$\frac{AD}{AB}\,=\,\frac{AB}{AC}$     ( Sides are proportional )

⇒ AD × AC = AB² ..............(1)

Now In ΔBDC and Δ ABC

∠C = ∠C  ( common angle )

∠BDC = ∠ABC ( right angle )

Thus, ΔBDC and Δ ABC are similar by AA similarity rule.

So,

$\frac{CD}{CB}\,=\,\frac{CB}{AC}$     ( Sides are proportional )

⇒ CD × AC = CB² ........... (2)

Add equation (1) & (2)

AD × AC + CD × AC = AB² + CB²

AC × ( AD + CD) = AB² + CB²

AC × AC = AB² + CB²

AC² = AB² + CB²

Hence Proved