in parallelogram (abcd), (mangle aeb = 112^circ), (mangle dca = 55^circ), (mangle dac = 81^circ), (ad = 7), (be = 8), and (ae = 5). find the measure of (angle dbc), (angle abd), and (angle dea).
note: the diagram is not drawn to scale.
(mangle dbc =)
(mangle abd =)
(mangle dea =)

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Accepted Answer

# Explanation:
## Step1: Find $ m\angle DBC $
In parallelogram $ABCD$, $AD \parallel BC$, so $ \angle DAC = \angle BCA = 81^\circ $ (alternate interior angles). In $\triangle BEC$, we know $ \angle DCA = 55^\circ $, but wait, actually, since $AD \parallel BC$, $ \angle DAC = \angle BCA = 81^\circ $? Wait, no, $ \angle DCA = 55^\circ $, and $AD \parallel BC$, so $ \angle DAC $ and $ \angle BCA $ are alternate interior angles? Wait, $AD \parallel BC$ and $AC$ is a transversal, so $ \angle DAC = \angle BCA = 81^\circ $? Wait, no, the given $ \angle DCA = 55^\circ $. Wait, maybe better to use triangle angle sum. Wait, in parallelogram $ABCD$, $AD = BC = 7$ (opposite sides of parallelogram are equal). Also, $AD \parallel BC$, so $ \angle DAC = \angle BCA = 81^\circ $ (alternate interior angles). Now, to find $ \angle DBC $, we can look at $\triangle BEC$? Wait, no, maybe first find $ \angle ADB $ or use the fact that $AB \parallel DC$, so $ \angle ABD = \angle CDB $. Wait, maybe start with $ \angle DBC $. Since $AD \parallel BC$, $ \angle DAC = \angle BCA = 81^\circ $ (alternate interior angles). Wait, but $ \angle DCA = 55^\circ $, and in triangle $ADC$, the sum of angles is $180^\circ$. Wait, $ \angle DAC = 81^\circ $, $ \angle DCA = 55^\circ $, so $ \angle ADC = 180 - 81 - 55 = 44^\circ $. Since $ABCD$ is a parallelogram, $ \angle ABC = \angle ADC = 44^\circ $? Wait, no, opposite angles of parallelogram are equal, so $ \angle ADC = \angle ABC $, and consecutive angles are supplementary. Wait, maybe I made a mistake. Let's correct: In $\triangle ADC$, angles are $ \angle DAC = 81^\circ $, $ \angle DCA = 55^\circ $, so $ \angle ADC = 180 - 81 - 55 = 44^\circ $. Then, since $AD \parallel BC$, $ \angle ADC + \angle BCD = 180^\circ $ (consecutive angles of parallelogram are supplementary), so $ \angle BCD = 180 - 44 = 136^\circ $. But $ \angle DCA = 55^\circ $, so $ \angle BCA = \angle BCD - \angle DCA = 136 - 55 = 81^\circ $, which matches the alternate interior angle from $AD \parallel BC$ (since $ \angle DAC = 81^\circ $ and $ \angle BCA = 81^\circ $, alternate interior angles, so $AD \parallel BC$ holds, which it does in a parallelogram). Now, to find $ \angle DBC $, we can use the fact that $AD \parallel BC$, so $ \angle ADB = \angle DBC $ (alternate interior angles, since $BD$ is a transversal). Wait, in $\triangle ADE$, we know $ \angle DAC = 81^\circ $, $AE = 5$, $AD = 7$, but maybe we can find $ \angle ADE $ first. Wait, no, let's look at $ \angle AEB = 112^\circ $, which is a vertical angle with $ \angle DEC $, so $ \angle DEC = 112^\circ $. Also, $ \angle AEB $ and $ \angle DEA $ are supplementary, so $ \angle DEA = 180 - 112 = 68^\circ $. Wait, that's a simpler step. Let's do that first for $ \angle DEA $.

## Step2: Find $ m\angle DEA $
$ \angle AEB $ and $ \angle DEA $ are adjacent angles forming a linear pair, so they are supplementary. Thus, $ m\angle DEA = 180^\circ - m\angle AEB $.
Given $ m\angle AEB = 112^\circ $, so $ m\angle DEA = 180 - 112 = 68^\circ $.

## Step3: Find $ m\angle DBC $
In parallelogram $ABCD$, $AD \parallel BC$, so $ \angle DAC = \angle BCA = 81^\circ $ (alternate interior angles, since $AC$ is a transversal). Now, consider $\triangle BEC$? Wait, no, let's look at $ \angle ADB $ and $ \angle DBC $. Since $AD \parallel BC$, $ \angle ADB = \angle DBC $ (alternate interior angles, $BD$ is transversal). To find $ \angle ADB $, we can use $\triangle ADE$. Wait, in $\triangle ADE$, we know $ \angle DAC = 81^\circ $, $ \angle DEA = 68^\circ $, so the sum of angles in a triangle is $180^\circ$, so $ \angle ADE = 180 - 81 - 68 = 31^\circ $. Wait, no, $ \angle DAC = 81^\circ $, $ \angle DEA = 68^\circ $, so $ \angle ADE = 180 - 81 - 68 = 31^\circ $. Then, since $ \angle ADE = \angle DBC $ (alternate interior angles, $AD \parallel BC$, $BD$ transversal), so $ m\angle DBC = 31^\circ $. Wait, but let's check again. Wait, $ \angle DAC = 81^\circ $, $ \angle DEA = 68^\circ $, so $ \angle ADE = 180 - 81 - 68 = 31^\circ $. Then, since $AD \parallel BC$, $ \angle ADE = \angle DBC $, so $ m\angle DBC = 31^\circ $.

## Step4: Find $ m\angle ABD $
In parallelogram $ABCD$, $AB \parallel DC$, so $ \angle ABD = \angle CDB $ (alternate interior angles, $BD$ is transversal). Also, in $\triangle ADC$, we found $ \angle ADC = 44^\circ $ (from step1: $180 - 81 - 55 = 44^\circ$). $ \angle ADC = \angle ADB + \angle CDB $. We found $ \angle ADB = 31^\circ $ (which is $ \angle ADE $), so $ \angle CDB = \angle ADC - \angle ADB = 44 - 31 = 13^\circ $? Wait, no, that can't be. Wait, maybe my mistake in step1. Let's recalculate $ \angle ADC $. Wait, $ \angle DAC = 81^\circ $, $ \angle DCA = 55^\circ $, so $ \angle ADC = 180 - 81 - 55 = 44^\circ $. Then, $ \angle ADB $ is part of $ \angle ADC $, so $ \angle ADB + \angle CDB = 44^\circ $. But we also know that in $\triangle ADE$, angles are $ \angle DAC = 81^\circ $, $ \angle DEA = 68^\circ $, so $ \angle ADE = 180 - 81 - 68 = 31^\circ $, which is $ \angle ADB $. Thus, $ \angle CDB = 44 - 31 = 13^\circ $. Since $AB \parallel DC$, $ \angle ABD = \angle CDB = 13^\circ $ (alternate interior angles).

Wait, let's verify with another approach. In parallelogram $ABCD$, $AB \parallel DC$, so $ \angle ABD = \angle CDB $. Also, $AD \parallel BC$, so $ \angle DBC = \angle ADB $. We found $ \angle ADB = 31^\circ $, so $ \angle DBC = 31^\circ $, and $ \angle CDB = 13^\circ $, so $ \angle ABD = 13^\circ $. Let's check the sum of angles in $\triangle ABD$: $ \angle ADB = 31^\circ $, $ \angle ABD = 13^\circ $, so $ \angle BAD = 180 - 31 - 13 = 136^\circ $. In parallelogram, $ \angle BAD + \angle ADC = 180^\circ $ (consecutive angles), so $ 136 + 44 = 180 $, which checks out. Good.

## Step5: Recap
- $ m\angle DEA = 68^\circ $ (linear pair with $ \angle AEB $)
- $ m\angle DBC = 31^\circ $ (alternate interior angle with $ \angle ADB $, which is found from $\triangle ADE$ angle sum)
- $ m\angle ABD = 13^\circ $ (alternate interior angle with $ \angle CDB $, which is $ \angle ADC - \angle ADB $)

Wait, let's recheck $ \angle DBC $. Since $AD \parallel BC$, $ \angle ADB = \angle DBC $ (alternate interior angles). In $\triangle ADE$, angles are $ \angle DAC = 81^\circ $, $ \angle DEA = 68^\circ $, so $ \angle ADE = 180 - 81 - 68 = 31^\circ $, so $ \angle DBC = 31^\circ $. Correct.

For $ \angle ABD $, since $AB \parallel DC$, $ \angle ABD = \angle CDB $. In $\triangle ADC$, $ \angle ADC = 180 - 81 - 55 = 44^\circ $, and $ \angle ADC = \angle ADB + \angle CDB $, so $ \angle CDB = 44 - 31 = 13^\circ $, so $ \angle ABD = 13^\circ $. Correct.

# Answer:
$ m\angle DBC = 31^\circ $
$ m\angle ABD = 13^\circ $
$ m\angle DEA = 68^\circ $

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QUESTION IMAGE

Question

Explanation:

Step1: Find \( m\angle DBC \)

Step2: Find \( m\angle DEA \)

Step3: Find \( m\angle DBC \)

Answer: