Question

Prove that in a right angles triangle the sqaue of the hypotenus is eqaul to the sum of the squares of the other two sides

Answer 1

GIVEN:

• A right-angled triangle ABC in which ∠B = 90°

TO PROVE:

• (Hypotenuse)² = (Base)² + (Perpendicular)² i.e., AC² = AB² + BC²

CONSTRUCTION:

From B draw BD ⊥ AC.

PROOF:

In triangles ADB and ABC, we have

$\rm{ \angle ADB = \angle ABC }$ [Each equal to 90°]

and,

$\rm{ \angle A = \angle A }$ [common]

So, by AA-similarity criterion, we have

$\rm{\triangle ADB \sim \triangle ABC}$

$\rm{\Rightarrow \dfrac{AD}{AB} = \dfrac{AB}{AC}}$ [ ∵ In similar triangles corresponding sides are proportional]

$\rm{\Rightarrow AB^2 = AD \times AC}$ .....1)

In triangles BDC and ABC, we have

$\rm{ \angle CDB = \angle ABC }$ [Each equal to 90°]

and,

$\rm{ \angle C = \angle C }$ [common]

So, by AA-similarity criterion, we have

$\rm{\triangle BDC \sim \triangle ABC}$

$\rm{\Rightarrow \dfrac{DC}{BC} = \dfrac{BC}{AC}}$ [ ∵ In similar triangles corresponding sides are proportional]

$\rm{\Rightarrow BC^2 = AC \times DC}$ .....2)

Adding. equations 1) and 2), we get

$\rm{ AB^2 + BC^2 = AD \times AC + AC \times DC}$

$\rm{\Rightarrow AB^2 + BC^2 = AC(AD + DC)}$

$\rm{\Rightarrow AB^2 + BC^2 = AC \times AC }$

Hence, AC² = AB² + BC²

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Answer 2

Prove that in a right angles triangle the sqaue of the hypotenus is eqaul to the sum of the squares of the other two sides