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in parallelogram abcd shown below, the bisectors of $angle abc$ and $an…

Question

in parallelogram abcd shown below, the bisectors of $angle abc$ and $angle dcb$ meet at e, a point on $overline{ad}$. if $mangle a = 68^circ$, determine and state $mangle bec$.

Explanation:

Step1: Find $\angle ABC$

In parallelogram $ABCD$, consecutive angles are supplementary.
$\angle ABC = 180^\circ - m\angle A = 180^\circ - 68^\circ = 112^\circ$

Step2: Find $\angle DCB$

Opposite angles in a parallelogram are equal.
$\angle DCB = m\angle A = 68^\circ$

Step3: Calculate bisected angles

BE bisects $\angle ABC$, so $\angle EBC = \frac{1}{2}\angle ABC = \frac{1}{2} \times 112^\circ = 56^\circ$
CE bisects $\angle DCB$, so $\angle ECB = \frac{1}{2}\angle DCB = \frac{1}{2} \times 68^\circ = 34^\circ$

Step4: Solve for $\angle BEC$

Sum of angles in $\triangle BEC$ is $180^\circ$.
$m\angle BEC = 180^\circ - \angle EBC - \angle ECB = 180^\circ - 56^\circ - 34^\circ$

Answer:

$90^\circ$