Question

if in ∆ ABC , angle 90°,than prove BC² =AB² + AC²
​

Answer 1

Given: In △ABC,m∠B=90

o

To prove: AC

2

=AB

2

+BC

2

Construction: Draw BD⊥AC

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.

△ADB∼△ABC

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

So,

AB

AD

=

AC

AB

or AD.AC=AB

2

....(1)

Also, △BDC∼△ABC

Similarly,

BC

CD

=

AC

BC

CD.AC=BC

2

...(2)

Adding (1) and (2)

AD.AC+CD.AC=AB

2

+BC

2

AC(AD+CD)=AB

2

+BC

2

AC.AC=AB

2

+BC

2

AC

2

=AB

2

+BC

2

Answer 2

This can be proved by phythagores theorem.

First let us draw a perpendicular BD from B to AC.

we get three right angled triangles : ABC, BCD, and ADB

Therefore,

$\frac{AB}{AD} = \frac{AC}{AB} \: \frac{AC}{BC} = \frac{BC}{CD}$

AB² = ACAD , BC²= AC.CD

Adding both, AB²+ BC²= AC.AD + AC.CD

AB²+ BC²= AC (AD + CD) Since AD+CD=AC

AB² + BC² = AC (AC)

AB²+ BC² = AC²

Hence proved