$CDE$ is an equilateral triangle formed on a side $CD$ of a square $ABCD$. Show that $\mathit{△}ADE\cong △BCE$.

In the given figure, if $AB=AC$ and $\angle B=\angle C$. Prove that $BQ=CP$.

If $△\mathrm{ABC}\cong △\mathrm{LKM}$, then side of $△\mathrm{LKM}$ equal to side $\mathrm{AC}$ of $△\mathrm{ABC}$ is

If $△\mathrm{ABC}\cong △\mathrm{ACB}$ then $△\mathrm{ABC}$ is isosceles with

If $△\mathrm{ABC}\cong △\mathrm{PQR}$ and $△\mathrm{ABC}$ is not congruent to $△\mathrm{RPQ}$, then which of the following is not true:

Which congruency condition applies in triangles $ABC$ and $PQR$ if the following conditions are true:

$AB=QP, \angle B=\angle P$ and $BC=PR$

In triangles $ABC$ and $PQR$, if $\angle A=\angle R, \angle B=\angle P$ and $AB=RP$, then which one of the following congruence conditions applies:

If $△PQR\cong △EFD$, then $ED=$