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

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

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

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

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

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

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

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

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

You are right, jashansinghai!
.

### content

Click on a term to search for related topics.