# Geometry question with congruent triangles and isosceles triangles

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?

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

$$\theta = \measuredangle ADE = \measuredangle AED = 42^{\circ} \tag{5}\label{eq5A}$$

• Thanks! I got it now :) . Sep 26, 2023 at 6:43
• @FutureMathperson You're welcome. Note I just added an updated version of your original diagram, as well as a few other details, to help make the answer a bit easier to follow. Sep 26, 2023 at 6:44
• That makes it even more clear. Thanks a lot! Sep 26, 2023 at 6:47

Here's a quicker solution that exploits the conditions $$AD = BC$$ and $$\alpha + \beta = 180^\circ$$ in a natural way.

A common trick is to construct $$F$$ such that $$\triangle BFC = \triangle ACD$$, which is possible because $$BC = AD$$.
Then, the $$\alpha + \beta = 180^\circ$$ condition implies that $$ACFB$$ is a cyclic quad.
Since $$AC = BF$$, hence this cyclic quad is isosceles trapezoidpezoid.
Thus $$42^\circ = \angle ABC = \angle BCF = \angle ADC = \theta$$.

You're really close!

For a solution using your hint and ideas that you already stated:

1. $$ABC$$ and $$CEA$$ are congruent triangles by SAS (by construction), which you guessed at.

2. Hence $$AD = BC = AE$$, giving us the isosceles triangle which you guessed at

3. Thus $$\angle ADC = \angle CEA = \angle ABC = 42^\circ$$