Question

4 the figure on the right is quadrilateral efgh formed by the intersection of the bisectors of the four angles in parallelogram abcd. answer the following questions.
(1) find the measure of \\( \angle bad + \angle abc \\).
(ans.)
(2) find the measure of \\( \angle eab + \angle abe \\).
(ans.)
(3) find the measure of \\( \angle hef \\).
(ans.)
(4) what kind of quadrilateral is efgh?
(ans.)

• remember bh || df, af || ch, and the result of 3.

Explanation:

Question 4 (1)

Step1: Recall parallelogram angle property

In a parallelogram, consecutive angles are supplementary (sum to \(180^\circ\)). \( \angle BAD \) and \( \angle ABC \) are consecutive angles of parallelogram \(ABCD\).

Step2: Apply the property

So, \( \angle BAD + \angle ABC = 180^\circ \).

Step1: Analyze angle bisectors

\(AE\) bisects \( \angle BAD \), so \( \angle EAB=\frac{1}{2}\angle BAD \). \(BE\) bisects \( \angle ABC \), so \( \angle ABE = \frac{1}{2}\angle ABC \).

Step2: Sum the angles

\( \angle EAB+\angle ABE=\frac{1}{2}(\angle BAD + \angle ABC) \). From part (1), \( \angle BAD+\angle ABC = 180^\circ \), so \( \frac{1}{2}\times180^\circ = 90^\circ \).

Step1: Recall triangle angle - sum and vertical angles

In \( \triangle ABE \), \( \angle AEB = 180^\circ-(\angle EAB + \angle ABE) \). From part (2), \( \angle EAB+\angle ABE = 90^\circ \), so \( \angle AEB=180 - 90=90^\circ \). \( \angle HEF \) and \( \angle AEB \) are vertical angles, so they are equal.

Step2: Conclude the measure

Thus, \( \angle HEF=\angle AEB = 90^\circ \).

Answer:

\(180^\circ\)

Question 4 (2)