1(xv)
In the given diagram, ∆ABC ∼ ∆PQR. If AD and PS are bisectors of ∠BAC and ∠QPR respectively then:
- (a)∆ABC ∼ ∆PQS
- (b)∆ABD ∼ ∆PQS
- (c)∆ABD ∼ ∆PSR
- (d)∆ABC ∼ ∆PSR
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Given,
∆ABC ∼ ∆PQR
∴ ∠A = ∠P
⇒ $\frac{∠A}{2} = \frac{∠P}{2}$
⇒ ∠BAD = ∠QPS
∠B = ∠Q [∵ ∆ABC ∼ ∆PQR]
In ∆ABD ∼ ∆PQS,
⇒ ∠BAD = ∠QPS
⇒ ∠B = ∠Q
∴ ∆ABD ∼ ∆PQS (By A.A. axiom)
Hence, Option 2 is the correct option.