Question

Given: 

In trapezium PQRS, side PQ || side SR, AR = 5 AP, AS = 5 AQ, then prove that SR = 5 PQ.

Answer 1

side PQ || side SR [Given] 

and seg SQ is their transversal.

∴ ∠QSR = ∠SQP [Altemate angles] 

∴ ∠ASR = ∠AQP (i) [Q – A – S]

In ∆ASR and ∆AQP,

∠ASR = ∠AQP [From (i)] 

∠SAR ≅ ∠QAP [Vertically opposite angles] 

∆ASR ~ ∆AQP [AA test of similarity]

\(\therefore\) \(\frac{AS}{AQ} = \frac{SR}{PQ}\) (ii) [Corresponding sides of similar triangles]

But, AS = 5 AQ [Given]

\(\therefore\) \(\frac{AS}{AQ} = \frac{5}{1}\) (iii)