In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.

In right triangle $ABC$,right-angled at $C$,$M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$ (see Fig). Show that:
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$

$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.

In the figure,$\angle B < \angle A$ and $\angle C < \angle D$. Show that $AD < BC$.

$D$ is a point on side $BC$ of $\Delta ABC$ such that $AD = AC$ (see figure). Show that $AB > AD$.

In the figure,sides $AB$ and $AC$ of $\Delta ABC$ are extended to points $P$ and $Q$ respectively. Also,$\angle PBC < \angle QCB$. Show that $AC > AB$.