Question

in the figure PQ||BC ,AQ =5cm, QC=10cm then prove that BC =3PQ​

Answer 1

Answer:

Step-by-step explanation:

We have

AB=AP+PB=(3+6)cm=9cm and,

AC=AQ+QC=(5+10)cm=15cm.

∴

AB

AP

​

=

9

3

​

=

3

1

​

and

AC

AQ

​

=

15

5

​

=

3

1

​

⇒

AB

AP

​

=

AC

AQ

​

Thus, in triangles APQ and ABC,we have

AB

AP

​

=

AC

AQ

​

and ∠A=∠A [Common]

Therefore, by SAS-criterion of similarity, we have

△APQ∼△ABC

⇒

AB

AP

​

=

BC

PQ

​

=

AC

AQ

​

⇒

BC

PQ

​

=

AC

AQ

​

⇒

BC

PQ

​

=

15

5

​

⇒

BC

PQ

​

=

3

1

​

⇒BC=3PQ [Hence proved