Question

P and Q are points on the sides AB and AC respectively of ABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3PQ

Answer 1

Given that AP=3cm
PB=6cm
AQ=5cm
QC=10cm

To show that BC=3PQ.

We can show this by using the converse of basic proportionality theorem.
AP/AB = AQ/AC
⇒3/3+6 =5/5+10
⇒3/9=5/15
⇒1/3=1/3 
 Since they are equal proportions, thus, PQ is parallel to BC
And,
AP/AB=AQ/AC=PQ/BC
⇒5/15=PQ/BC
⇒1/3=PQ/BC
∴BC=3PQ 
Hence proved

Answer 2

Given : AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm.

To prove : BC = 3 PQ

Proof : AP/PB = 3/6 = 1/2

→ AQ/QC = 5/10 = 1/2

This concludes that, AP/PB = AQ/QC.

Therefore, PQ || BC. Hence ang APQ = ang ABC and ang AQP = ang ACB. [Corresponding angles]

Thus, by AA similarity criteria, ∆APQ ~ ∆ABC.

Hence, AP/AB = PQ/BC

→ 3/(3 + 6) = PQ/BC

→ 1/3 = PQ/BC

→ BC = 3 PQ

Q.E.D