Question

$△ABC\mathrm{and}△DBC$ are two isosceles triangles on the same base $BC.$ Show that $\angle \mathrm{ABD}=\angle \mathrm{ACD}$.

Answer 1

Proving $\angle \mathrm{ABD}=\angle \mathrm{ACD}$:

In $∆\mathrm{ABC}$,

$\mathrm{AB}=\mathrm{AC}$ ($∆\mathrm{ABC}$ is isosceles triangle)

Therefore, $\angle \mathrm{ABC}=\angle \mathrm{ACB}$ [angles opposite to equal sides are equal]…………$\left(1\right)$

In $∆\mathrm{DBC}$,

$\mathrm{DB}=\mathrm{DC}$ ($∆\mathrm{DBC}$ is isosceles triangle)

Therefore, $\angle \mathrm{DBC}=\angle \mathrm{DCB}$ [angles opposite to equal sides are equal]………..$\left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$, we get

$\begin{array}{rcl}\angle \mathrm{ABC}+\angle \mathrm{DBC}& =& \angle \mathrm{ACB}+\angle \mathrm{DCB}\\ \angle \mathrm{ABD}& =& \angle \mathrm{ACD}\end{array}$

Hence Proved