Determine the value, in degrees, of x in the quadrilateral below

Question

In the figure below, $AB = BC$. Determine the value, in degrees, of x ($Answer:10^o$)

The solution I found was

$3x = x+20 \implies x = 10^o$

How to prove that the $\angle DBE=20^o$?

Answer 1

You almost have it, but we can somewhat more easily finish by just handling things in the opposite order (although, as I indicate below, there are actually $2$ possible answers). Similar to what you've shown, draw $BE$ just so that $\measuredangle DBE=20^{\circ}$. Then we also have $\measuredangle DAB = \measuredangle DEB = 160^{\circ} - x$, which means the $3$ angles in $\triangle DAB$ match those in $\triangle DEB$. Thus, since $DB$ is a common side, its side length is the same for both triangles, so we get

$$\triangle DAB \cong \triangle DEB \;\;\to\;\; \lvert AB \rvert = \lvert BE \rvert$$

With it being given that $\lvert AB \rvert = \lvert BC \rvert$, this means $\lvert BE \rvert = \lvert BC \rvert$. However, next note that, based on just the provided information, without assuming the figure is necessarily a particularly accurate representation, there are now actually $3$ possibilities to check (although just something like stating $\measuredangle DBC$ is $\gt 20^{\circ}$ or that it's obtuse, would have been sufficient for there to be only the first possibility below).

First, if $E$ is to the left of $C$ (i.e., in $DC$, as shown in your figure), then $\triangle BEC$ is isosceles. Thus, by also using your results, we then have that

$$\measuredangle BEC = 3x \;\;\to\;\; 3x = x + 20^{\circ} \;\;\to\;\; x = 10^{\circ}$$

The second possibility is that $E$ is actually the same point as $C$. In this case, we get from the angles in $\triangle BED$ that

$$\measuredangle BED = 3x \;\;\to\;\; x + 3x + 20^{\circ} = 180^{\circ} \;\;\to\;\; x = 40^{\circ}$$

The final, third possibility to consider is that $E$ is to the right of $C$, i.e., those $2$ points are basically switched around in your diagram. We would then have $\measuredangle BCE = \measuredangle BEC = 180^{\circ}-3x$. Then using the sum of angles in $\triangle DEB$ gives that

$$x + (180^{\circ}-3x) + 20^{\circ} = 180^{\circ} \;\;\to\;\; -2x + 20^{\circ} = 0 \;\;\to\;\; x = 10^{\circ}$$

On the other hand, since $0^{\circ} \lt \measuredangle CBE \lt 20^{\circ}$, we get from $\triangle CBE$ that

$$\begin{equation}\begin{aligned} 180^{\circ} & \gt 2(180^{\circ}-3x) \gt 160^{\circ} \\ 180^{\circ} & \gt 360^{\circ}-6x \gt 160^{\circ} \\ -180^{\circ} & \gt -6x \gt -200^{\circ} \\ 30^{\circ} & \lt x \lt \frac{100}{3}^{\circ} \\ \end{aligned}\end{equation}$$

However, this contradicts the previous equation's result, so this third possibility is not a valid case.

Answer 2

The cicumdiameter of $\triangle BAD$ is $\frac{AB}{\sin(x)}$. The cicumdiameter of $\triangle BCD$ is $\frac{BC}{\sin(x)}$. These are equal since $AB=BC$.

Since $BD$ is a chord of both the circumcircle of $\triangle BAD$ and the circumcircle of $\triangle BCD$, we get that either $\angle BAD=\angle BCD$ or $\angle BAD=180^{\large\circ}-\angle BCD$. That is, there are two solutions.

---

Solution 1: $\angle BAD=\color{#C00}{\angle BCD}$.

Since the angles in $\triangle BAD$ sum to $180^{\large\circ}$, $$ x+\color{#C00}{3x}+20^{\large\circ}=180^{\large\circ}\implies x=40^{\large\circ}\tag1 $$

$\triangle BAD$ and $\triangle BCD$ are both $20^{\large\circ}{-}120^{\large\circ}{-}40^{\large\circ}$.

---

Solution 2: $\angle BAD=\color{#C00}{180^{\large\circ}-\angle BCD}$.

Since the angles in $\triangle BAD$ sum to $180^{\large\circ}$, $$ x+\color{#C00}{180^{\large\circ}-3x}+20^{\large\circ}=180^{\large\circ}\implies x=10^{\large\circ}\tag2 $$

$\triangle BAD$ is $20^{\large\circ}{-}150^{\large\circ}{-}10^{\large\circ}$ and $\triangle BCD$ is $140^{\large\circ}{-}30^{\large\circ}{-}10^{\large\circ}$.

Solution 2 is the one found in the question.