Question

If APB and CQD are two parallel lines, then the bisectors of $\angle APQ,\angle BPQ,\angle CQP$ and $\angle PQD$ forms

• A. a square
• B. a rhombus
• C. a rectangle
• D. any other parallelogram

Answer 1

The correct option is C a rectangle

$\angle APB+\angle BPQ={180}^o$ ...angles in linear pair

$\therefore \angle APE+\angle QPE+\angle BPF+\angle FPQ={180}^o$

$\therefore 2\angle EPQ+2\angle FPQ={180}^o$

[Since PE and PF are bisectors of $\angle APQ$ and $\angle BPQ$ respectively]

Dividing by 2 on both sides we get,

$\angle EPQ+\angle FPQ={90}^o$

$\therefore \angle EPF={90}^o$ ...angle addition property ...(I)

Similarly, we can prove $\angle EQF={90}^o$ ...(II)

Now, $APB\parallel CQD$ ...(given)

$\therefore \angle APQ\cong \angle PQD$ ...alternate angles

and $\angle BPQ\cong \angle CQP$ ...alternate angles

$\therefore \frac{1}{2}\angle APQ=\frac{1}{2}\angle PQD$ and $\frac{1}{2}\angle BPQ=\frac{1}{2}\angle CQP$

$\therefore \angle EPQ=\angle FQP$ and $\angle EQP=\angle FPQ$ ...(III)

In $△EPQ$ and $△FPQ$,

$\therefore \angle EPQ\cong \angle FQP$ ...from (III)

$\angle EQP\cong \angle FPQ$ ...from (III)

side $PQ\cong$ side $PQ$ ... common side

$\therefore △EPQ\cong △FPQ$ ...by $ASA$ test

$\therefore \angle E\cong \angle F$ ....C.P.C.T ....(IV)

In $\square EPFQ$

$\angle E+\angle EPF+\angle F+\angle EQF={360}^o$ ...Angle sum property of quadrilateral

$\therefore 2\angle E+{90}^o+{90}^o={360}^o$ ...from (I), (II), (IV)

$\therefore 2\angle E={180}^o$

$\therefore \angle E={90}^o$

$\therefore \angle E=\angle F={90}^o$ ...from (IV) ...(V)

$\therefore \square EPFQ$ is a rectangle ...from (I), (II) (V)

[By definition of rectangle]