Question

$△\mathrm{ABC}$ is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{BD}$ and $\mathrm{CE}$ are its two medians. Show that$\mathrm{BD}=\mathrm{CE}$.

Answer 1

Showing $BD=CE$:

Given:

$\text{ABC}$ is an isosceles triangle with $\mathrm{AB}=\mathrm{AC}$

$\mathrm{BD}$ and $\mathrm{CE}$ are its two medians

To Prove:

$\mathrm{BD}=\mathrm{CE}.$

Proof:

In $\mathrm{\Delta ABD}$ and $\mathrm{\Delta ACE}$,

$\mathrm{AB}=\mathrm{AC}$ (given)

$2\mathrm{AE}=2\mathrm{AD}$ (as D and E are mid-points)

So $\mathrm{AE}=\mathrm{AD}$

$\angle \mathrm{A}=\angle \mathrm{A}$ (common)

Hence,$\mathrm{\Delta ABD}\cong \mathrm{\Delta ACE}$ (By $\mathrm{SAS}$ criteria )

Therefore $BD=CE$[ by $\mathrm{CPCT}$]

Hence we proved that $\mathrm{BD}=\mathrm{CE}$