Question

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.)

1. find the measure of \\( \angle eab + \angle abe \\).

(ans.)

1. find the measure of \\( \angle hef \\).

(ans.)

1. what kind of quadrilateral is efgh?

(ans.)
★ remember \\( bh \parallel df \\), \\( af \parallel ch \\), and the result of 3.

Explanation:

1)

Step1: Recall parallelogram property

In a parallelogram \(ABCD\), consecutive angles are supplementary, i.e., \(\angle BAD + \angle ABC = 180^\circ\) (since \(AD \parallel BC\), same - side interior angles are supplementary).

Step1: Analyze angle bisectors

Since \(AE\) bisects \(\angle BAD\) and \(BE\) bisects \(\angle ABC\), we have \(\angle EAB=\frac{1}{2}\angle BAD\) and \(\angle ABE = \frac{1}{2}\angle ABC\).

Step2: Calculate the sum

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

Step1: Recall triangle angle - sum and vertical angles

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

Answer:

\(180^\circ\)

2)