# In Fig. 6.36 QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR

**Solution:**

We know if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

This is referred to as SAS (Side - Angle - Side) similarity criterion for two triangles.

In ΔPQR,

∠1 = ∠2

⇒ PR = PQ (In a triangle sides opposite to equal angles are equal)

In ΔPQS and ΔTQR

∠PQS = ∠TQR = ∠1 (same angle)

QR/QS = QT/PQ (Since PR = PQ)

⇒ ΔPQS ~ ΔTQR (SAS criterion)

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 6

**Video Solution:**

## In Fig. 6.36 QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.3 Question 4

**Summary:**

Hence it is proved that ΔPQS ~ ΔTQR if, QR/QS = QT/PR and ∠1 = ∠2 in the given figure.

**☛ Related Questions:**

- S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
- In Figure 6.37, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
- In Fig. 6.38, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC.
- E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that ΔABE ~ ΔCFB.