In the diagram below $AD\equiv BC$ and $\alpha + \beta=180^{\circ}$. Find the measure of $\theta$.
I'm given a hint:
First extend $DC$ past $C$ to the point $E$ where $CE \equiv AB$, then draw $AE$ and look for congruent triangles.
So I did extend the figure and I get something like this:
Now $\Delta ADE$ ALMOST looks like an isosceles triangle which would imply $\theta = 42^{\circ}$ but I'm not sure how to justify this.
It also looks like $\Delta ACE$ and $\Delta ABC$ are congruent which would imply $\angle AEC = 42^{\circ}$ but again, I can't be sure.
Can someone help out?
Best Answer
You've made a good start. As shown in the updated diagram below, draw a line from $E$ perpendicular to $AC$, meeting it at $F$, and similarly from $B$ perpendicular to $AC$, meeting it at $G$. Also, join $B$ to $E$ and $A$ to $E$.
Note that
$$\measuredangle ACE = \measuredangle CAB = \alpha, \;\;\; \lvert CE\rvert = \lvert AB\rvert \tag{1}\label{eq1A}$$
We then get that $\measuredangle CEF = \measuredangle ABG$, so
$$\triangle CEF \cong \triangle ABG \;\;\to\;\; \lvert EF \rvert = \lvert BG \rvert \tag{2}\label{eq2A}$$
Thus, $BEFG$ is a rectangle, so $BE \parallel AC$, which means $ABEC$ is an isosceles trapezoid. Thus, as stated in that article, the opposite angles are supplementary, so it's also a cyclic quadrilateral, and therefore as the subtended angles on the same side as $AC$ are equal, we get
$$\measuredangle ABC = \measuredangle AEC = 42^{\circ} \tag{3}\label{eq3A}$$
Since isosceles trapezoid diagonal lengths are of equal length, then
$$\lvert AE\rvert = \lvert BC\rvert = \lvert AD\rvert \tag{4}\label{eq4A}$$
Thus, as you suggested, $\triangle ADE$ is an isosceles triangle, so we get from \eqref{eq3A} that
$$\theta = \measuredangle ADE = \measuredangle AED = 42^{\circ} \tag{5}\label{eq5A}$$