# Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆ PQR

**Solution:**

As 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 as SAS criterion for two triangles.

Produce AD to E so that AD = DE. Join CE

Similarly, produce PM to N such that PM = MN , and Join RN

In ΔABD and ΔCDE

AD = DE [By Construction]

BD = DC [ AP is the median]

∠ADB = ∠CDE [Vertically opposite angles]

Therefore, ΔABD ≅ ΔCDE [By SAS criterion of congruence]

⇒ AB = CE. [CPCT] ...(i)

Also, in ΔPQM and DMNR

PM = MN. [By Construction]

QM = MR [ PM is the median]

∠PMQ = ∠NMR

Therefore, ΔPQM = ΔMNR

[Vertically opposite angles]

[By SAS criterion of congruence]

⇒ PQ = RN [CPCT]...(ii)

Now,

AB / PQ = AC / PR = AD / PM [Given]

⇒ CE / RN = AC / PR = AD / PM [from (i) and (ii)]

⇒ CE / RN = AC / PR = 2AD / 2PM

⇒ CE / RN = AC / PR = AE / PN [ 2AD = AE and 2PM = PN ]

Therefore, ΔACE ~ ΔPRN [By SSS similarity criterion]

Therefore, ∠CAE = ∠RPN

Similarly, ∠BAE = ∠QPN

Hence, ∠CAE + ∠BAE = ∠RPN + ∠QPN

⇒ ∠BAC = ∠QPR

⇒ ∠A = ∠P ....(iii)

Now, In ΔABC and ΔPQR

AB/PQ = AC/PR

∠A = ∠P [from (iii)]

Therefore, ΔABC ~ ΔPQR [By SAS similarity criterion]

**Video Solution:**

## Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR

### NCERT Class 10 Maths - Chapter 6 Exercise 6.3 Question 14:

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC~ ΔPQR

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Hence proved that ΔABC~ ΔPQR