Question

Prove that if in a triangle, the square on one side is equal to the sum of the squares on the other two sides, then the angle opposite to the first side is a right angle.

Answer 1

FIGURE IS IN THE ATTACHMENT.

SOLUTION:

GIVEN:
A ∆ABC , in which AC² = AB² + BC²…….(1)

To Prove : ∠B = 90°

Construction : construct a ∆PQR right angled at Q , Such that
PQ = AB & QR = BC ……….(2)

PROOF:
In ∆PQR ,
PR² = PQ² + QR²
[By Pythagoras theorem]

PR² = AB² + BC²……………..(3)
[From equation 2]

From eq 1 & 3,
AC² = PR²
AC = PR……………………..(4)

[Taking square root on both sides]

In ∆ABC & ∆PQR,
AB = PQ   [From equation 2]
BC = QR   [From equation 2]
AC = PR   [From equation 4]
∆ABC ≅  ∆PQR

[By SSS Congruence rule]

∠B = ∠Q   [CPCT]
∠Q = 90°   [By construction]
∠B = ∠Q  = 90°

Hence, ∠B = 90°

This theorem is converse of Pythagoras theorem.

HOPE THIS WILL HELP YOU...

Answer 2

➡ Given :- In ∆ABC, BC^2 = AB^2 + AC^2.

➡ To Prove :- angle A = 90°

➡ Proof :- Let ray OX be a ray.

Construct ray OY such that ray OY perpendicular to ray OX.

Let M belongs to ray OY such that OM = AC
Let N belongs to ray OX such that ON = AB
Draw MN.

∆OMN is a right angled triangle, as OM perpendicular to ON

angle MON is right angle.

MN is the hypotenuse.

According to Pythagoras theorem.

MN^2 = OM^ 2 + ON^2
MN^2 = AC^2 + AB^2

But AB^2 + AC^2 = BC^2

MN^2 = BC^2
MN = BC. .... ( 1 )

In ∆ABC and ∆ONM consider the correspondence ABC ↔ONM, we have
AB = ON
AC = OM
BC = MN

The correspondence ABC ↔ ONM is a congruence. so, ∆ABC = ∆ONM ... ( by SSS the. )

angle A = angle O

But angle O in ∆ONM is right angle by construction.

angle A is a right angle. ( proved )

This theorem is called Converse of Pythagoras Theorem.