3155.41 – A Mystery Angle

In the figure, triangle $\triangle ABC$ has $CA = CB$ and angle $\angle C = 20^\circ$ . In addition, $BD = DC$ and $BE = BA$ . What is angle $\angle BDE$ ?

Detailed Hint: Draw the arc of a circle with center $B$ and radius $BD$ meeting $BA$ extended at $X$ and $BC$ at $Y$ . Draw segments $DX$ and $DY$ .

Solution

The figure shows the arc and lines suggested by the hint. To find angle BDE = D_3, we find D_2, D_4, and D_5. Then subtract from 180 degrees.

1. $\triangle ABC$ is isosceles, therefore $A_1 = 80^\circ$ . $\triangle BDC$ is isosceles, therefore $B_2 = 20^\circ$ and $B_1 = 60^\circ$ . Note also that $A_2 = 100^\circ$ .

2. $\triangle BDY$ is isosceles, therefore $Y_1 = D_3 + D_4 = 80^\circ$ . So $D_5 = 60^\circ$ . Note that $Y_2 is now$ 100^\circ$.

3. $\triangle XBD$ is isosceles with $B_1 = 60^\circ$ , therefore $\triangle XBD$ is equilateral, hence $\angle X = 60^\circ$ and $XD = BD = DC$ .

4. $\angle X = D_5$ , $XD = BD = DC$ , and $D_1 = \angle C = 20^\circ$ , therefore $\triangle XAD \cong \triangle DYC$ by ASA.

5. From 4, we get $XA = DY$ .

6. Since $BX - BA = BY - BE$ , $XA = EY$ . Thus $EY = DY$ and $\triangle DYE$ is isosceles.

7. We now conclude that $E_2 = D_4 = 50^\circ$ .

8. In $\triangle BAD$ ,

$\begin{align*}D_2 &= 180^\circ - B_1 - A_1 \\&= 180^\circ - 60^\circ - 80^\circ \\&= 40^\circ\end{align*}$

9. $\begin{align*}D_3 &= 180^\circ - D_2 - D_4 - D_5 \\&= 180^\circ - 40^\circ - 50^\circ - 60^\circ \\&= 30^\circ\end{align*}$

This is the answer!