# If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding side and the median of another triangle, the prove that the two triangles are similar.

## Solution :

Given : Two triangles ABC and DEF, in which AP and DQ are the medians, such that $$AB\over DE$$ = $$AC\over DF$$ = $$AP\over DQ$$

To prove : $$\triangle$$ ABC ~ $$\triangle$$ DEF

Construction : Produce line AP to G, so that PG = AP. Join CG. And Produce line DQ to H, so that QH = DQ. Join FH.

Proof : In triangles APB and GPC, BP = CP            (because AP is the median)

AP = GP        (by construction)

and    $$\angle$$ APB = \$$\angle$$ CPG       (vertically opposite angles)

$$\therefore$$  By SAS theorem of congruence,

$$\triangle$$ APB $$\cong$$ $$\triangle$$ GPC

$$\implies$$  AB = GC             …….(1)           (C.P.C.T)

Again, in triangles DQE and HQF,

EQ = FQ                 (because DQ is the median)

DQ = HQ                (by construction)

and  $$\angle$$ DQE = $$\angle$$ HQF       (vertically opposite angles)

$$\therefore$$  By SAS theorem of congruence,

$$\triangle$$ DQE $$\cong$$ $$\triangle$$ HQF

$$\implies$$  DE = HF          ………(2)        (C.P.C.T)

Now,  $$AB\over DE$$ = $$AC\over DF$$ = $$AP\over DQ$$            (given)

From (1) and (2),     AB = GC  and   DE = HF

$$\implies$$  $$GC\over HF$$ = $$AC\over DF$$ = $$AP\over DQ$$

$$\implies$$ $$GC\over HF$$ = $$AC\over DF$$ = $$2AP\over 2DQ$$

$$\implies$$  $$GC\over HF$$ = $$AC\over DF$$ = $$AG\over DH$$

By SSS similarity,

$$\triangle$$ AGC ~ $$\triangle$$ DHF

$$\implies$$  $$\angle$$ 1 = $$\angle$$ 2

Similarly, $$\angle$$ 3 = $$\angle$$ 4

Thus,  $$\angle$$ 1 + $$\angle$$ 3 = $$\angle$$ 2 + $$\angle$$ 4

$$\implies$$  $$\angle$$ A = $$\angle$$ D         ……….(3)

Thus, in triangles ABC and DEF,

$$\angle$$ A = $$\angle$$ D               (from 3)

and $$AB\over DE$$ = $$AC\over DF$$         (given)

So, By SAS similarity,

$$\triangle$$ ABC ~ $$\triangle$$ DEF