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# S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS

**Solution:**

Let's draw ∆PQR as per the given question.

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred to as the AA similarity criterion for two triangles.

In ΔRPQ and ΔRTS,

∠RPQ = ∠RTS (given)

∠PRQ = ∠TRS (common angle)

Thus, ΔRPQ ∼ ΔRTS (AA criterion)

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

**Video Solution:**

## S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.3 Question 5

**Summary:**

If S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS, we proved that ΔRPQ ~ ΔRTS using AA criteria.

**☛ Related Questions:**

- 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.
- In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP.

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