Question

If two medians of a triangle are equal then prove that it is an isosceles triangle

Answer 1

Proved below.

Step-by-step explanation:

Given:

Let ABC be a triangle.

Taking A as origin, let position vectors of B and C be b and c respectively.

So, the position vectors of the mid-points FF and EE of sides AB and AC are $\frac{b}{2}$ and $\frac{c}{2}$ respectively.

∴ $BE = \frac{b}{2}-b, CF = \frac{c}{2}-c$                 [1]

Now, BE=CF (given),

∴ $BE ^2 = CF ^2$

⇒ $(\frac{c}{2}-b )^2=(\frac{b}{2}-c )^2$                    [ from 1 ]  

⇒ $\frac{c^2}{4}+b^2-2\times \frac{c}{2}\times b=\frac{b^2}{4}+c^2-2\times\frac{b}{2}\times c$

⇒ $\frac{c^2}{4}+b^2-c\times b=\frac{b^2}{4}+c^2 -b\times c$

⇒ $\frac{c^2+4b^2}{4} =\frac{b^2+4c^2}{4}$

⇒ $\frac{c^2+4b^2}{4} -[\frac{b^2+4c^2}{4}]=0$

⇒ $\frac{c^2+4b^2-b^2-4c^2}{4}=0$

⇒ $\frac{3b^2-3c^2}{4} =0$

⇒ $\frac{3}{4}(b^2-c^2) =0$

⇒ $b^2-c^2=0\times \frac{3}{4}$

⇒ $b^2-c^2=0$

⇒ $b^2=c^2$

⇒ $AB^2=AC^2$

⇒ AB = AC

Hence the triangle is isosceles.

Hence proved.