Question

In a given figure,AP is bisector of angle a and CQ is the bisector of angle c of parallelogram ABCD. Prove that APCQ is a parallelogram

Answer 1

APCQ is a parallelogram

Step-by-step explanation:

Parallelogram has opposite sides Parallel & equal in length

and opposite angles in parallelogram are equal

∠A = ∠C

=> Bisector of ∠A & ∠C are equal

in ΔAPB  & ΔDCQ

AB = CD  ( opposite sides of parallelogram)

∠ABD = ∠CDP  (BD intersecting two parallel line)

∠BAP = ∠DCQ  (  bisectors of ∠A & ∠C)

=> ΔAPB  ≅ ΔDCQ

=> AP = CQ

BP = DQ

BQ = BP + PQ

DP = DQ + PQ

=> BQ = DP

now in ΔABQ  & ΔCDP

AB = CD  ( opposite sides of parallelogram)

BQ = DP  ( shown above)

∠ABQ = ∠CDP  

ΔABQ  ≅ ΔCDP

=> AQ = CP

AP = CQ , CP = AQ

hence APCQ is a parallelogram

Answer 2

APCQ is a parallelogram.

Step-by-step explanation:

We are given a figure in which AP is the bisector of angle A and CQ is the bisector of angle C of parallelogram ABCD.

We have to prove that ABCD is a parallelogram.

Firstly, we have to make construction in the figure by joining A to C.

Now, as it is given that;

$\angle BAP = \frac{1}{2} \angle A$     {$\because$ AP is the bisector of angle A}

and, $\angle DCQ = \frac{1}{2} \angle C$       {$\because$ CQ is the bisector of angle C}

Since we know that $\angle$A and $\angle$C are equal because opposite angles of the parallelogram (ABCD) are equal. This means;

$\angle$BAP = $\angle$DCQ   ----------- [Equation 1]

Now, as ABCD is a parallelogram and $AB || CD$ ;

So, $\angle$BAC = $\angle$DCA     {aternate angles are equal} ------- [Equation 2]

Subtracting equation 1 and 2 we get;

$\angle$BAP - $\angle$BAC = $\angle$DCQ - $\angle$DCA

$\angle$CAP = $\angle$ACQ

This means that $AP || QC$ as alternate angles are equal.

Similarly, we can show that $AQ|| PC$.

Hence, APCQ is a parallelogram.