Question

Q.57. In a triangle if the square on one side is equal to the sum of
squares on the other two sides, prove that the angle opposite
to the first side is a right angle.
Use the above theorem to prove the following:
In a quadrilateral ABCD (Fig. 34) angle B = 90°. IF AD2 = AB2
+ BC+ CD2, prove that angle ACD = 90°​

Answer 1

Answer:

Given:- ABC is a triangle

AC

2

=AB

2

+BC

2

To prove:- ∠B=90°

Construction:- Construct a triangle PQR right angled at Q such that, PQ=AB and QR=BC

Proof:-

In △PQR

PR

2

=PQ

2

+QR

2

(By pythagoras theorem)

⇒PR

2

=AB

2

+BC

2

.....(1)(∵AB=PQ and QR=BC)

AC

2

=AB

2

+BC

2

.....(2)(Given)

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

AC

2

=PR

2

⇒AC=PR.....(3)

Now, in △ABC and △PQR

AB=PQ

BC=QR

AC=PR(From (3))

∴△ABC≅△PQR(By SSS congruency)

Therefore, by C.P.C.T.,

∠B=∠Q

∵∠Q=90°

∴∠B=90°

Hence proved.