V K Raman and S N Sharma Solutions for Exercise 2: EXERCISE 24(B)

In the figure, $\mathrm{BD}=\mathrm{DC}, \mathrm{DE}\perp \mathrm{AB}, \mathrm{DF}\perp \mathrm{AC}$ and $\mathrm{DE}=\mathrm{DF}$. Prove that $\angle \mathrm{ABD} =\angle \mathrm{ACD}$ .

In the rectangle $\mathrm{ABCD}$, prove that $\angle \mathrm{BAC}=\angle \mathrm{BDC}$ .

$\mathrm{ABCD}$ is a parallelogram. Prove that $∆\mathrm{ABD}\equiv ∆\mathrm{CDB}$.

$\mathrm{OD}$ is in the interior of $\angle \mathrm{AOB}$ . $\mathrm{P}$ is any point in $\mathrm{OD}, \mathrm{PC}, \mathrm{PD}$ are perpendicular on $\mathrm{OA},\mathrm{OB}$ respectively. If $\mathrm{PC}=\mathrm{PD}$, prove that$\angle \mathrm{POC}=\angle \mathrm{POD}$ .

$\mathrm{ABC}$ is an isosceles $∆$ . $\mathrm{BCDE}$ is a square. $\mathrm{AE}, \mathrm{AD}$ are joined. Prove that$∆\mathrm{ABE}\equiv ∆\mathrm{ACD}$ .

In the quadrilateral $\mathrm{ABCD}, \mathrm{AC}$ bisects $\angle \mathrm{A}$ and $\angle \mathrm{C}$. Prove that $\mathrm{AB}=\mathrm{AD}$ and $\mathrm{BC}=\mathrm{DC}$.

Show that each angle of an equilateral triangle is $60°$ .

Find the value of the acute angles in an isosceles right angled triangle.