Question

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

Answer 1

Given: In △ABC and △DEF, AM and DN are medians to BC and EF, respectively, and

`(AB)/(DE) = (AC)/(DF) = (AM)/(AN)`

Let, `(AB)/(DE) = (AC)/(DF) = (AM)/(AN)` = k

So, AB = k DE, AC = k DF, AM = k DN

Now, in △ABC, since AM is a median to BC, by Apollonius theorem,

AB² + AC² = 2(AM² + BM²)

But `BM = (BC)/2`, so

`AB^2 + AC^2 = 2(AM^2 + (BC^2)/4)`

`AB^2 + AC^2 = 2AM^2 + (BC^2)/2`

2AB² + 2AC² = 4AM² + BC²

BC² = 2AB² + 2AC² − 4AM²    ...(1)

Similarly, in △DEF, since DN is a median to EF,

EF² = 2DE² + 2DF² − 4DN²    ...(2)

Substituting AB = k DE, AC = k DF, AM = k DN in (1), we get:

BC² = 2(k DE)² + 2(k DF)² − 4(k DN)²

BC² = k²(2DE² + 2DF² − 4DN²)

Using equation (2),

BC² = k²EF²

BC = k EF

`(BC)/(EF) = k`

But already,

`(AB)/(DE) =  k, (AC)/(DF) = k`

Therefore,

`(AB)/(DE) = (AC)/(DF) = (BC)/(EF)`

So, the three sides of △ABC are proportional to the three corresponding sides of △DEF.

∴ △ABC ∼ △DEF    ...(by SSS similarity criterion)

Hence proved.