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 thattriangles

\(AB\over DE\) = \(AC\over DF\) = \(AP\over DQ\)

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

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