Question

in the diagram below, ( overline{ab}paralleloverline{cd} ), ( overline{ad}paralleloverline{bc} ), ( mangle eab = 27^{circ} ) and ( mangle dea = 60^{circ} ). find ( mangle cde ).

Explanation:

Step1: Identify triangle angle sum

In $\triangle ADE$, the sum of angles is $180^\circ$. We know $\angle DEA = 60^\circ$ and $\angle EAD = 27^\circ$ (since $AD \parallel BC$ and $AB \parallel CD$, $ABCD$ is a parallelogram, so $\angle EAB = \angle ECD$? Wait, no, first find $\angle ADE$.
Using angle sum: $m\angle ADE = 180^\circ - m\angle DEA - m\angle EAD$.
Substitute values: $m\angle ADE = 180^\circ - 60^\circ - 27^\circ = 93^\circ$? Wait, no, wait. Wait, $\angle EAB$ is $27^\circ$, but $\angle EAD$—wait, maybe I misread. Wait, the diagram: $\angle EAB$ is $27^\circ$? Wait, the label at $A$: $\angle BAE$ is $27^\circ$? Wait, no, the angle at $A$ between $AB$ and $AE$ is $27^\circ$? Wait, no, the problem says $m\angle EAB = 27^\circ$ and $m\angle DEA = 60^\circ$. Wait, $\triangle ADE$: angles are $\angle DEA = 60^\circ$, $\angle EAD$—wait, $AB \parallel CD$, so $\angle EAB = \angle EDC$ (alternate interior angles). Wait, maybe better: since $AB \parallel CD$ and $AD \parallel BC$, $ABCD$ is a parallelogram, so $AB \parallel CD$, so $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles). Then in $\triangle CDE$? Wait, no, $\angle DEA$ is $60^\circ$, so its vertical angle $\angle BEC$ is $60^\circ$, but maybe not. Wait, let's correct:

Wait, $\triangle ADE$: angles are $\angle DEA = 60^\circ$ (given), $\angle EAD$—wait, the angle at $A$: $\angle EAB = 27^\circ$, but $AD$ is a side, so $\angle EAD$ is part of $\angle DAB$. Wait, maybe I made a mistake. Let's start over.

Given $AB \parallel CD$ and $AD \parallel BC$, so $ABCD$ is a parallelogram. Therefore, $AB \parallel CD$, so $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles, transversal $AC$? Wait, no, transversal $AD$? Wait, $AB \parallel CD$, transversal $AD$: $\angle BAD + \angle CDA = 180^\circ$, but maybe not. Wait, the problem is to find $m\angle CDE$. Let's look at $\triangle ADE$:

In $\triangle ADE$, $m\angle DEA = 60^\circ$, $m\angle EAD$—wait, the angle at $A$: $\angle EAB = 27^\circ$, but $AE$ is a diagonal. Wait, maybe $\angle EAD$ is equal to $\angle EBC$? No, better: use triangle angle sum in $\triangle ADE$. Wait, $\angle DEA = 60^\circ$, $\angle EAD$: wait, the problem says $m\angle EAB = 27^\circ$, so $\angle EAD$—wait, maybe $\angle EAD$ is not $27^\circ$. Wait, no, the problem states $m\angle EAB = 27^\circ$ and $m\angle DEA = 60^\circ$. Let's assume that in $\triangle ADE$, angles are $\angle DEA = 60^\circ$, $\angle EAD$ (let's call it $x$), and $\angle ADE$ (let's call it $y$). But we also know that $AB \parallel CD$, so $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles, transversal $AD$? No, transversal $AE$? Wait, $AB \parallel CD$, so $\angle BAE = \angle CDE$ (alternate interior angles, transversal $AD$? No, $AB$ and $CD$ are parallel, $AD$ is a transversal, so $\angle BAD + \angle CDA = 180^\circ$, but maybe not. Wait, maybe the key is that $\angle DEA$ is $60^\circ$, so in $\triangle ADE$, the third angle is $180 - 60 - 27 = 93^\circ$? No, that can't be. Wait, no, $\angle EAB$ is $27^\circ$, but $\angle EAD$ is not $27^\circ$. Wait, I think I misread the diagram: the angle at $A$ between $AB$ and $AE$ is $27^\circ$, and $\angle DEA$ is $60^\circ$. Then in $\triangle ADE$, $\angle EAD = 27^\circ$, $\angle DEA = 60^\circ$, so $\angle ADE = 180 - 60 - 27 = 93^\circ$? But that seems off. Wait, no, maybe $\angle CDE$ is equal to $\angle ADE$? No, wait, $AB \parallel CD$, so $\angle EAB = \angle EDC = 27^\circ$, and then in $\triangle CDE$, we need another angle. Wait, maybe the vertical angle: $\angle DEA = 60^\circ$, so $\angle BEC = 60^\circ$, but maybe not. Wait, the problem is to find $m\angle CDE$. Let's try again:

Since $AB \parallel CD$, $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles, transversal $AC$? No, transversal $AE$? Wait, $AB \parallel CD$, so when transversal $AD$ cuts them, $\angle BAD + \angle CDA = 180^\circ$, but maybe not. Wait, the correct approach: in $\triangle ADE$, angles sum to $180^\circ$. So $m\angle ADE = 180^\circ - m\angle DEA - m\angle EAD$. Wait, $m\angle EAD$: the problem says $m\angle EAB = 27^\circ$, so if $AE$ is a diagonal, then $\angle EAD$ is part of $\angle DAB$, but maybe $\angle EAD = \angle EAB$? No, that doesn't make sense. Wait, maybe the diagram shows that $\angle BAE = 27^\circ$, and since $AB \parallel CD$, $\angle CDE = \angle BAE = 27^\circ$? No, that can't be, because $\angle DEA$ is $60^\circ$. Wait, I think I made a mistake. Let's check the angle sum in $\triangle ADE$ again.

Wait, $\angle DEA = 60^\circ$ (given), $\angle EAD = 27^\circ$ (given as $\angle EAB$? Wait, maybe $\angle EAB$ is $\angle EAD$? No, the label is $\angle EAB$ (between $AB$ and $AE$), so $AE$ is a diagonal, so $AB$ and $AE$ form $27^\circ$, and $AD$ is another side. Wait, maybe the correct way is: since $AB \parallel CD$, $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles, transversal $AC$? No, transversal $AD$? Wait, $AB \parallel CD$, transversal $AD$: $\angle BAD + \angle CDA = 180^\circ$, but $AD \parallel BC$, so $ABCD$ is a parallelogram, so $AB = CD$, $AD = BC$, and $AB \parallel CD$, $AD \parallel BC$. Therefore, $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles, transversal $AC$? No, transversal $AE$? Wait, $AB \parallel CD$, so $\angle BAE = \angle DCE$ (alternate interior angles, transversal $AC$), but that's not helpful. Wait, maybe the triangle $\triangle ADE$: angles are $\angle DEA = 60^\circ$, $\angle EAD = 27^\circ$, so $\angle ADE = 180 - 60 - 27 = 93^\circ$? No, that's too big. Wait, no, maybe $\angle EAD$ is not $27^\circ$. Wait, the problem says $m\angle EAB = 27^\circ$ and $m\angle DEA = 60^\circ$. So in $\triangle ADE$, the angles are $\angle DEA = 60^\circ$, $\angle EAD$ (let's say $x$), and $\angle ADE$ (let's say $y$). Then $x + y + 60 = 180$, so $x + y = 120$. But since $AB \parallel CD$, $\angle EAB = \angle EDC = 27^\circ$ (alternate interior angles, transversal $AD$), so $y = 27^\circ$? No, that would mean $x = 93^\circ$, which is possible. Wait, but the question is to find $m\angle CDE$, which is $\angle EDC$, which is equal to $\angle EAB = 27^\circ$? No, that can't be, because $\angle DEA$ is $60^\circ$. Wait, I'm confused. Wait, maybe the correct answer is $33^\circ$? Wait, no, let's do it properly.

Wait, $\triangle ADE$: $\angle DEA = 60^\circ$, $\angle EAD = 27^\circ$, so $\angle ADE = 180 - 60 - 27 = 93^\circ$? No, that's not right. Wait, maybe $\angle EAB$ is $27^\circ$, and $\angle DEA$ is $60^\circ$, so in $\triangle ABE$, but no. Wait, the key is that $AB \parallel CD$, so $\angle CDE = \angle AED$? No, alternate interior angles. Wait, I think I made a mistake in the angle identification. Let's start over:

Given:
- $AB \parallel CD$
- $AD \parallel BC$
- $m\angle EAB = 27^\circ$
- $m\angle DEA = 60^\circ$

We need to find $m\angle CDE$.

Since $AB \parallel CD$, $\angle EAB$ and $\angle EDC$ are alternate interior angles (transversal $AD$? No, transversal $AE$? Wait, $AB \parallel CD$, transversal $AE$: $\angle EAB$ and $\angle ECD$ are alternate interior angles. No, transversal $AD$: $\angle BAD$ and $\angle CDA$ are supplementary. Wait, maybe $\triangle ADE$:

Angles in $\triangle ADE$:
- $\angle DEA = 60^\circ$ (given)
- $\angle EAD$: let's say this is equal to $\angle EAB = 27^\circ$ (if $AE$ is a bisector, but not necessarily). Wait, no, the diagram shows $\angle BAE = 27^\circ$, so $AE$ is a diagonal, so $\angle EAD$ is another angle. Wait, maybe the correct approach is:

In $\triangle ADE$, sum of angles is $180^\circ$. So:
$m\angle ADE = 180^\circ - m\angle DEA - m\angle EAD$

But $m\angle EAD = m\angle EAB = 27^\circ$ (assuming $AE$ is along $AD$? No, that's not possible. Wait, I think the mistake is here: $\angle EAB$ is $27^\circ$, and since $AB \parallel CD$, $\angle CDE = \angle AED - \angle EAB$? No, that's not a theorem. Wait, maybe the answer is $33^\circ$? Wait, $60 - 27 = 33$? No, angle sum. Wait, no, let's calculate again.

Wait, $\angle DEA = 60^\circ$ (vertex angle), $\angle EAB = 27^\circ$ (alternate interior angle with $\angle EDC$). Wait, no, $\angle DEA$ is an exterior angle to $\triangle CDE$? No, $\angle DEA$ is equal to $\angle EDC + \angle ECD$ (exterior angle theorem). But if $AB \parallel CD$, $\angle EAB = \angle ECD = 27^\circ$ (alternate interior angles). Then $\angle DEA = \angle EDC + \angle ECD$, so $60^\circ = m\angle CDE + 27^\circ$, so $m\angle CDE = 60^\circ - 27^\circ = 33^\circ$. Ah! That makes sense. Exterior angle theorem: in $\triangle CDE$, $\angle DEA$ is an exterior angle, so it's equal to the sum of the two non-adjacent interior angles ($\angle EDC$ and $\angle ECD$). Since $AB \parallel CD$, $\angle EAB = \angle ECD = 27^\circ$ (alternate interior angles, transversal $AC$). Therefore, $m\angle DEA = m\angle CDE + m\angle ECD$, so $60^\circ = m\angle CDE + 27^\circ$, so $m\angle CDE = 60^\circ - 27^\circ = 33^\circ$.

Step2: Apply exterior angle theorem

In $\triangle CDE$, $\angle DEA$ is an exterior angle, so $m\angle DEA = m\angle CDE + m\angle DCE$.
Since $AB \parallel CD$, $m\angle DCE = m\angle EAB = 27^\circ$ (alternate interior angles).
Substitute: $60^\circ = m\angle CDE + 27^\circ$.
Solve for $m\angle CDE$: $m\angle CDE = 60^\circ - 27^\circ = 33^\circ$.

Answer:

$33^\circ$