In an isosceles triangle, prove that the altitude from the vertex bisects the base.

In the given figure, if $AB=AC$, then prove that $\angle ABD=\angle ACE$.

In the given figure, if $BC=2.6 \mathrm{cm}$, then find $2BD+\frac{BC}{2}$.

In the given figure, if $l\Vert m$, $\angle ABC=\angle ABD=40°$ and $\angle BAC=\angle BAD=90°$, then prove that $∆BCD$ is an isosceles triangle.

In the given figure, $ABCD$ is a square and $P$ is the mid-point of $AD . BP$ and $CP$ are joined. Prove that $\angle PCB=\angle PBC$.

In the figure, the diagonal $AC$ of a quadrilateral $ABCD$ bisects $\angle A$ and $\angle C$. Prove that $AB=AD$ and $CB=CD$.

In the given figure, $\angle B=\angle E, BD=CE$ and $\angle 1=\angle 2$. Show that $∆ABC\cong ∆AED$.