Question

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides​

Answer 1

$\sf \huge {Given:-}$

--》Right triangle ABC right angled at B

$\sf \huge{To \: prove:-}$

--》AC² = AB² + BC²

$\sf \huge{Solution:-}$

--》Draw BD perpendicular to AC

Now,

--》ADB ~ ABC

(ADB~ABC ----》 If a perpendicular is drawn from the vertex if the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.)

--》AD/AB = AB/ AC (Sides are proportional)

(Or)

--》AD . AC = AB² (1)

Also,

--》BDC ~ ABC (Above mentioned)

--》CD/BC = BC/AC

(Or)

--》CD.AC = BC² (2)

Adding 1 and 2

--》AD .AC + CD .AC = AB²+ BC²

--》AC (AD+CD) = AB² + BC²

--》AC. AC = AB² + BC²

--》AC² = AB² + BC²

HENCE PROVED

Answer 2

Given:-

A right triangle ABC right angled at B.

To prove:-

AC² =AB ² +BC²

Construction:-

Draw BD⊥AC

Proof:-

In △ABC and △ABD

∠ABC=∠ADB(Each 90°)

∠A=∠A(Common)

∴△ABC∼△ABD(By AA)

$\frac{AC}{AB}$

= $\frac{AB}{AD}$

(∵Sides of similar triangles are proportional)

⇒AB²

=AD⋅AC.....(1)

Similarly, in △ABC and △BCD

∠ABC=∠BDC(Each 90°)

∠C=∠C(Common)

∴△ABC∼△BCD(By AA)

∴ $\frac{BC}{DC}$

= $\frac{AC}{BC}$

⇒BC²

=DC⋅AC.....(2)

Adding equation (1)&(2), we have

AB² +BC²

=AD⋅AC+DC.AC

⇒AB² +BC²

=AC(AD+DC)

⇒AB²+BC²

=AC²

Hence proved.