Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle 

In an equilateral triangle with side $a$, prove that area $=\frac{\sqrt{3}}{4}{a}^2$.

Find the length of each side of a rhombus whose diagonals are $24 \mathrm{cm}$ and $10 \mathrm{cm}$ long.

Prove that the ratio of the perimeters of two similar triangles is the ratio of their corresponding sides.

In the given figure, $△ABC$ and $△DBC$ have the base $BC$. If $AD$ and $BC$ intersect at $O$, Prove that $\frac{ar\left(△ABC\right)}{ar\left(△DBC\right)}=\frac{AO}{DO}$.

In the given figure, $XY\parallel AC$ and $XY$ divides $△ABC$ into two regions, equal in area. Show that $\frac{AX}{AB}=\frac{\left(2-\sqrt{2}\right)}{2}$.

In the given figure, $△ABC$ is an obtuse triangle, obtuse-angled at $B$. If $AD\perp CB$ (produced) prove that $A{C}^2=A{B}^2+B{C}^2+2BC.BD$.

In the given figure, each one of $PA, QB$ and $RC$ is perpendicular to $AC$. If $AP=x, QB=z, RC=y, AB=a$ and $BC = b$ show that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$.