In a $\triangle ABC$ median $AD$ is produced to $X$ such that $AD = DX$. Prove that $ABXC$ is a parallelogram.
Given:
In a $\triangle ABC$ median $AD$ is produced to $X$ such that $AD = DX$.
To do:
We have to prove that $ABXC$ is a parallelogram.
Solution:
Join $BX$ and $CX$
In $\triangle ABD$ and $\triangle CDX$,
$AD = DX$
$BD = DC$
$\angle ADB = \angle CDX$ (Vertically opposite angles)
Therefore, by SAD axiom,
$\triangle ABD \cong \triangle CDX$
This implies,
$AB = CX$ (CPCT)
$\triangle ABD = \triangle DCX$
$\triangle ABD$ and $\triangle DCX$e are alternate angles.
$AB \parallel CX$ and $AB = CX$
Therefore,
$ABXC$ is a parallelogram.
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