Question

In the following figure,ABCD is a parallelogram.
(i) $APbisects\angle A$
(ii) $BPbisects\angle B$

then the statements $1,2$ are

• A. True
• B. False

Answer 1

The correct option is A True

ABCD is parallelogram $AD=DP=PC=CB$

To check

(i) AP bisect angle A

(ii) BP bisects angle B

(iii) $\angle DAP-\angle CBP=\angle APS$

$\Rightarrow$ (i) ABCD is parallelogram

$AB||DC\Rightarrow AB||DP$ and $AB||PC$.

Now, $In\Delta ADP$

$AD=DP$

$\therefore \angle DPA=\angle DAP$

$\angle 3=\angle 1$ __(1) [Angles opposite to equal sides of a triangle are equal]

also, $\angle 3=\angle 2$ __(2) [AP acts as transversal for $AB||DP$]

From (1) and (2) [So, alternate interior angles are equal]

$\therefore AP$ bisects angle $\angle DAB$

(ii) In $\Delta PCB$,

$CB=PC$

$\angle 4=\angle 5$ __(3) [Angles opposite to equal sides of a triangle are equal]

Also, $PC||AB$

$\therefore PB$ acts as transversal.

Alternate interior angles are equal.

$\angle 4=\angle 6$ ___(4)

From (3) and (4),

$\angle 5=\angle 6$

$\therefore PB$ bisects angle B.