Question

ΔPQR is isosceles with PQ = QR. QR is extended to S so that ΔPRS becomes isosceles with PR = PS. Show that ∠PSRSP : ∠QPS = 1 : 3

Answer 1

In ΔPQR,
PQ = QR    ...(given)
∠PRQ = ∠QPR   .....(i)
In ΔPRS,
PR = RS    ...(given)
∠PSR = ∠RPS   .......(ii)
Adding (i) and (ii)
∠QPR + ∠RPS = ∠PRQ + ∠PSR
∠QPS = ∠PRQ + ∠PSR   .........(iii)
Now in PRS,
∠PRQ = ∠RPS + ∠PSR
∠PRQ = ∠PSR + ∠PSR    ...(from (ii))
∠PRQ = 2∠PSR   ..............(iv)
Now, ∠QPS = 2∠PSR + ∠PSR  ...(from (iii) and (iv))
∠QPS = 3∠PSR
`(∠"PSR")/(∠"QPS") = (1)/(3)`
⇒ ∠PSR = ∠QPS = 1 : 3.