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 Δ ABC ~ Δ PQR.

Δ ABC and Δ PQR are shown in the figure below:

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

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

In Δ ABD and Δ CDE,

⇒ AD = DE [By Construction]

⇒ BD = DC [AD is the median]

⇒ ∠ADB = ∠CDE [Vertically opposite angles are equzl]

∴ Δ ABD ≅ Δ ECD [By S.A.S. axiom]

⇒ AB = CE [C.P.C.T.] ………..(1)

Also, in Δ PQM and Δ MNR,

⇒ PM = MN [By Construction]

⇒ QM = MR [PM is the median]

⇒ ∠PMQ = ∠NMR [Vertically opposite angles are equal]

∴ Δ PQM ≅ Δ NRM [By SAS axiom]

⇒ PQ = RN [C.P.C.T.]…………(2)

Now,

$\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$ [Given]

⇒ $\dfrac{CE}{RN} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$ [from (1) and (2)]

⇒ $\dfrac{CE}{RN} = \dfrac{AC}{PR} = \dfrac{2AD}{2PM}$

⇒ $\dfrac{CE}{RN} = \dfrac{AC}{PR} = \dfrac{AE}{PN}$ [ 2AD = AE and 2PM = PN ]

∴ Δ ACE ~ Δ PRN [By SSS similarity criterion]

∴ ∠CAE = ∠RPN

Similarly, ∠BAE = ∠QPN

Hence, ∠CAE + ∠BAE = ∠RPN + ∠QPN

⇒ ∠BAC = ∠QPR

⇒ ∠A = ∠P ….(2)

Now, In Δ ABC and Δ PQR,

$\dfrac{AB}{PQ} = \dfrac{AC}{PR}$

∠A = ∠P [from (2)]

∴ Δ ABC ~ Δ PQR [By SAS similarity criterion]

Hence, Δ ABC ~ Δ PQR.