In the diagram below $AD\equiv BC$ and $\alpha + \beta=180^{\circ}$. Find the measure of $\theta$.

enter image description here

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:

enter image description here

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?


3 Answers 3


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 $E$.

Diagram of OP with lines EF and BG drawn perpendicular to AC, plus BE and AE added

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

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

  • $\begingroup$ Thanks! I got it now :) . $\endgroup$ Sep 26 at 6:43
  • $\begingroup$ @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. $\endgroup$ Sep 26 at 6:44
  • 1
    $\begingroup$ That makes it even more clear. Thanks a lot! $\endgroup$ Sep 26 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 $\angle ADC = \angle CEA = \angle ABC = 42^\circ$


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