Question

in the following figure, \\(\overline{eb}\\) and \\(\overline{dc}\\) intersect at point a.
what is the value of \\(x\\) in the figure?
\\(x = \square\\)

Explanation:

Step1: Prove triangle congruence

We can see that $EA = CA = 4.2$, $DA = BA = 4$, and $\angle EAD = \angle CAB$ (vertical angles are equal). By the SAS (Side-Angle-Side) congruence criterion, $\triangle EAD \cong \triangle CAB$.

Step2: Find $\angle B$

In $\triangle ABC$, the sum of interior angles is $180^\circ$. Calculate $\angle B$:
$\angle B = 180^\circ - 36^\circ - 56^\circ = 88^\circ$

Step3: Match congruent angles

Since $\triangle EAD \cong \triangle CAB$, $\angle D = \angle B = 88^\circ$.

Step4: Identify vertical angle pair

$\angle x$ and $\angle EAD$ are supplementary? No, $\angle x$ is equal to the exterior angle of $\triangle ABC$ or we use the fact that $\angle x$ is equal to the angle corresponding to $\angle BAC$'s opposite? Wait, no: $\angle x$ is equal to the angle at $B$ plus angle at $C$? No, wait, $\angle x$ is vertical to the angle that is supplementary to $\angle EAD$? No, correction: $\angle x$ is equal to the measure of $\angle ABC + \angle ACB$? No, wait, $\angle x$ is equal to the angle that is corresponding to the angle in $\triangle ABC$: actually, $\angle x$ is equal to $180^\circ - \angle EAD$, but $\angle EAD = \angle CAB$? No, wait, no: $\angle x$ is the angle at $A$ in $\triangle EAD$, which is equal to $\angle CAB$? No, no, vertical angles: $\angle x$ and the angle adjacent to $36^\circ$ are vertical? No, wait, $\overline{EB}$ and $\overline{DC}$ intersect at $A$, so $\angle x$ and the angle opposite to it (the angle at $A$ in $\triangle ABC$ that is not $36^\circ$) is not. Wait, no, let's re-express:

Wait, $\triangle EAD \cong \triangle CAB$, so $\angle EAD = \angle CAB$? No, $\angle EAD$ is $\angle x$, and $\angle CAB$ is $36^\circ$? No, no, the intersection: $\angle x$ and $\angle CAB$ are vertical angles? No, $\angle x$ is $\angle EAD$, and $\angle CAB$ is the angle at $A$ in $\triangle ABC$, which is $36^\circ$. Wait, no, vertical angles are $\angle EAC$ and $\angle DAB$, $\angle x$ is $\angle EAD$, which is supplementary to $\angle CAB$? No, no, let's use the sum of angles in $\triangle ABC$:

In $\triangle ABC$, $\angle CAB = 36^\circ$, $\angle ACB = 56^\circ$, so $\angle ABC = 180 - 36 - 56 = 88^\circ$.

Since $\triangle EAD \cong \triangle CAB$, $\angle ADE = \angle ABC = 88^\circ$, $\angle AED = \angle ACB = 56^\circ$, so in $\triangle EAD$, $\angle x = 180 - 88 - 56 = 36^\circ$? No, that can't be. Wait, no, the vertical angles: $\angle x$ is equal to the angle that is the exterior angle of $\triangle ABC$? No, wait, $\angle x$ is equal to $\angle ABC + \angle ACB$? No, that's $88+56=144$? No, no, I messed up the congruence.

Wait, $EA = CA = 4.2$, $AD = AB = 4$, and $\angle EAD = \angle CAB$ (vertical angles). So $\triangle EAD \cong \triangle CAB$ (SAS). Therefore, $\angle E = \angle C = 56^\circ$, $\angle D = \angle B = 88^\circ$. Then in $\triangle EAD$, $\angle x = 180 - 56 - 88 = 36^\circ$? No, but $\angle x$ and $\angle CAB$ are vertical angles, so they should be equal. Oh right! $\angle x$ and $\angle CAB$ are vertical angles? No, $\overline{EB}$ and $\overline{DC}$ intersect at $A$, so $\angle EAD$ (which is $\angle x$) and $\angle CAB$ are vertical angles, so they are equal. Wait, but $\angle CAB$ is $36^\circ$? But that contradicts the sum. Wait no, looking at the figure: $\angle CAB$ is $36^\circ$, $\angle ACB$ is $56^\circ$, so $\angle ABC = 180 - 36 - 56 = 88^\circ$. Then $\triangle EAD$ has sides $EA=4.2$, $AD=4$, same as $CA=4.2$, $AB=4$, so $\triangle EAD \cong \triangle CAB$, so $\angle EAD = \angle CAB = 36^\circ$? But that would make $x=36$, but that can't be, because $\angle x$ is labeled as $\angle EAD$, which is vertical to $\angle CAB$. Wait, no, the figure shows $\angle x$ is $\angle EAD$, and $\angle CAB$ is $36^\circ$, so they are vertical angles, so $x=36$? But that's too easy. Wait no, no, the figure: $\overline{EB}$ and $\overline{DC}$ intersect at $A$, so $\angle EAC$ and $\angle DAB$ are vertical angles, $\angle EAD$ and $\angle CAB$ are vertical angles. So $\angle x = \angle CAB = 36^\circ$? But why are the triangles congruent? Because $EA=CA$, $AD=AB$, and vertical angles, so SAS, so $\triangle EAD \cong \triangle CAB$, so $\angle x = \angle CAB = 36^\circ$. Wait, but that's the same as vertical angles. Oh right, vertical angles are equal, so $x=36$? But wait, no, the sum in $\triangle ABC$ is 180, which checks out, and congruent triangles confirm it.

Wait, no, I made a mistake earlier: $\angle x$ is $\angle EAD$, which is vertical to $\angle CAB$, which is $36^\circ$, so $x=36$. But let's confirm with the triangle sum: in $\triangle EAD$, if $x=36$, then $\angle E + \angle D + 36 = 180$, so $\angle E + \angle D = 144$. Since $\triangle EAD \cong \triangle CAB$, $\angle E = \angle C = 56$, $\angle D = \angle B = 88$, $56+88=144$, which is $180-36=144$, so that's correct.

Wait, but why did I overcomplicate? Vertical angles are equal, so $x=36$. But let's re-express the steps correctly:

Step1: Identify vertical angles

$\angle x$ and $\angle CAB$ are vertical angles.

Step2: State vertical angle property

Vertical angles are congruent, so $\angle x = \angle CAB$.

Step3: Assign given value

$\angle CAB = 36^\circ$, so $x=36$.

Wait, but the congruent triangles are just confirmation. The simplest way is vertical angles.

Answer:

$x=36$