Question

In ΔPQR, QS is the bisector of ∠PQR. S lies on PR. T is any point on QS such that a line through T parallel to PQ intersect QR at K, then ΔQKT is always


An isosceles triangle


An equilateral triangle


A right-angled triangle


An isosceles right-angled triangle

Answer 1

Answer:

We have,

According to given figure.

PQ=PR(giventhat)

QS=SR(Bydefinationofmidpoint)

PS=PS(Commonline)

Then,

ΔSPQ≅ΔSPR (BY congruency S.S.S.)

Hence, PS bisects ∠PQR by definition of angle bisector.

Answer 2

Given, PS⊥QR

Let PS=2x,QS=4x,RS=x

Now, in △PSR

PR

2

=PS

2

+RS

2

PR

2

=4x

2

+x

2

PR=

5

x

In △PQS

PQ

2

=PS

2

+QS

2

PQ

2

=4x

2

+16x

2

PQ=

20

x

Now, PQ

2

+PR

2

=20x

2

+5x

2

=25x

2

and, PS

2

=(4x+x)

2

=25x

2

Thus, PQ

2

+PR

2

=PS

2

By converse of Pythagoras theorem, the triangle is a right angled triangle.