$\mathrm{D}$ is a point on the base $\mathrm{BC}$ of an isosceles triangle $\mathrm{ABC}$ in which $\mathrm{AB}=\mathrm{AC}$. The triangle $\mathrm{ADE}$ is drawn such that $\mathrm{AD}=\mathrm{AE}$, $\angle \mathrm{DAE}=\angle \mathrm{BAC}$ and $\mathrm{D},\mathrm{E}$ are opposite sides of $\mathrm{AC}$. Prove that $\mathrm{AC}$ bisects $\angle \mathrm{BCE}$.

In the given figure, which is not drawn accurately, $\angle \mathrm{BAD}=59°, \angle \mathrm{DAC}=32°$ and $\mathrm{AD}=\mathrm{BD}$. Calculate the value of the angle $\mathrm{ACB}$ and state, giving reasons, which is greater, $\mathrm{BD} \mathrm{or} \mathrm{DC}$.

In given figure, $\mathrm{ABCD}$ is an isosceles trapezium. Find the values of $x$ and $y$.

Given, $\mathrm{AC}=5x+3$ and $\mathrm{BD}=9x-12$, find the value of $x$ so that $\mathrm{ABCD}$ is isosceles. (Express answer in decimal)

Find $\mathrm{SR}$.

In this figure, $\mathrm{ABCD}$ is a kite whose diagonals intersect at $\mathrm{O}$. If $\angle \mathrm{BAD}=110°$ and $\angle \mathrm{BCD}=50°$, find $\angle \mathrm{OBC}$.

In this figure, $\mathrm{ABCD}$ is a kite whose diagonals intersect at $\mathrm{O}$. If $\angle \mathrm{BAD}=110°$ and $\angle \mathrm{BCD}=50°$, find $\angle \mathrm{ODA}$.