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# Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR

**Solution:**

We know that 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 similarity criterion for two triangles.

In ΔABC and ΔPQR

AB/PQ = BC/QR = AD/PM [given]

AD and PM are median of ΔABC and ΔPQR respectively

⇒ BD/QM = (BC/2)/(QR/2) = BC/QR

Now, in ΔABD and ΔPQM

AB/PQ = BD/QM = AD/PM

⇒ ΔABD ∼ ΔPQM [SSS criterion]

Now, in ΔABC and ΔPQR

AB/PQ = BC/QR [given in the statement]

∠ABC = ∠PQR [∵ ΔABD ∼ ΔPQM]

⇒ ΔABC ∼ ΔPQR [SAS criteion]

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

**Video Solution:**

## Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Fig. 6.41). Show that ∆ ABC ~ ∆ PQR

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.3 Question 12

**Summary:**

In the above figure, sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ, QR, and median PM of ∆PQR. Hence proved that ∆ABC ~ ∆ PQR.

**☛ Related Questions:**

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