Question

A transversal cuts $2$ parallel lines $PQ$ and $RS$ at point $A$ and $B$ respectively. The two interior angles at $A$ are bisected ans so are the two interior angle at $B$, the four bisectors from a quadrilateral $ABCD$, prove that :
(i) $ABCD$ is a rectangle (ii) $CD$ is parallel to given parallel

Answer 1

$\begin{array}{c}\angle PAB=\angle \left(AH\right)P\\ \frac{1}{2}\angle PAB=\frac{1}{2}\angle ABS\\ \angle CAB=\angle ABD\\ \left(bi\sec torofA,\angle B\right)\\ alternatingangles1ptareequalso,\\ AC\parallel BD\\ similarly,BC\parallel AD\\ bothpairsparalleltoso,ABCDisparalle\log ram\\ forPAQ\\ \angle PAB+\angle QAB={180}^0\left(linearpair\right)\\ \frac{1}{2}\angle PAB+\frac{1}{2}QAB={90}^0\\ \angle CAB+\angle DAB={90}^0\\ \angle CAD={90}^0\\ inaparalle\log ram\angle CAD={90}^{0}so,\\ itisarec\tan gle.\end{array}$