1. Finding an angle

Triangle ABC where AB=AC. D is on AC such that BC=BD, and E is on AB where BE=ED=AD.

2. is the answer 45 degrees

3. Hello, Denny!

Triangle $\displaystyle ABC$ where $\displaystyle AB=AC$.
$\displaystyle D$ is on $\displaystyle AC$ such that $\displaystyle BC=BD$.
$\displaystyle E$ is on $\displaystyle AB$ such that $\displaystyle BE=ED=AD$.

What is angle A?
Code:
                  A
o
/ \
/   \
/     \ D
/       o
/   *  *  \
E o     *     \
/    *        \
/   *           \
/  *              \
/ *                 \
/*                    \
B o - - - - - - - - - - - o C
Let: $\displaystyle \alpha \:=\:\angle A$

$\displaystyle \Delta ABC$ is isosceles: .$\displaystyle \angle B = \angle C$
Then: .$\displaystyle \angle A + \angle B + \angle C \:=\:180^o \quad\Rightarrow\quad \alpha + 2\angle C \:=\:180^o \quad\Rightarrow\quad \angle C\:=\:\frac{180^o-\alpha}{2}$

Since $\displaystyle BC = BD,\:\Delta BCD$ is isosceles: .$\displaystyle \angle BDC \:=\:\angle C \quad\Rightarrow\quad \angle BDC \:=\:\frac{180^o-\alpha}{2}\;{\color{blue}[1]}$

Since $\displaystyle ED = AD,\:\Delta ADE$ is isosceles: .$\displaystyle \angle AED \:=\:\angle A \:=\:\alpha$
. . Then: .$\displaystyle \angle ADE \:=\:180^o - 2\alpha\;{\color{blue}[2]}\:\text{ and }\:\angle BED \:=\:180^o-\alpha$

Since $\displaystyle BE = ED,\:\Delta BED$ is isosceles: .$\displaystyle \angle EBD = \angle EDB$
. . $\displaystyle \angle BED \;+\; \angle EBD \;+\; \angle EDB \:=\:180^o \quad\Rightarrow\quad(180^o-\alpha) \;+\; 2\angle EDB \:=\:180^o$ . $\displaystyle \Rightarrow\quad \angle EDB \:=\:\frac{\alpha}{2}\;{\color{blue}[3]}$

At point $\displaystyle D$, we have: .$\displaystyle \angle ADE + \angle EDB + \angle BDC \:=\:180^o$

Substitute $\displaystyle {\color{blue}[2]}, {\color{blue}[3]}, {\color{blue}[1]}$: .$\displaystyle (180^o - 2\alpha) + \frac{\alpha}{2} + \frac{180^o-\alpha}{2} \:=\:180^o$

Solve for $\displaystyle \alpha\!:\;\;\boxed{\alpha \:=\:45^o}$

You are right, jashansinghai!
.