In Fig. 6.36, 
and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.
Given: In ΔPQR,
∠PQR = ∠PRQ
PQ = PR (i)
Given,
Using (i),
(ii)
In triangle PQS and TQR, we get
[Using (i)]
∠Q = ∠Q
Therefore,
Triangle PQS 
TQR (SAS similarity rule)
22
In Fig. 6.36, and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.
Given: In ΔPQR,
∠PQR = ∠PRQ
PQ = PR (i)
Given,
Using (i),
(ii)
In triangle PQS and TQR, we get
[Using (i)]
∠Q = ∠Q
Therefore,
Triangle PQS TQR (SAS similarity rule)