Question

The bisectors of the angle of a parallelogram enclose a

Options

• parallelogram

•  rhombus

•  rectangle

• ^square

Answer 1

We have ABCD, a parallelogram given below:

Therefore, we have AB || BC

Now,  AD || BC and transversal AB intersects them at A and B respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

∠A + ∠B = 180°

`1/2∠A + 1/2∠B = 90°`

We have AR and BR as bisectors of ∠A and ∠B  respectively.

 ∠RAB +∠RBA = 90° …… (i)

Now, in ΔABR, by angle sum property of a triangle, we get:

∠RAB + ∠RBA +∠ARB = 180° 

From equation (i), we get:

90° + ∠ARB = 180°

∠ARB = 90° 

Similarly, we can prove that ∠DPC = 90° .

Now, AB || DC and transversal ADintersects them at A and D respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

∠A + ∠D = 180°

`1/2∠A+1/2 ∠D = 90°` 

We have AR and DP as bisectors of ∠A and ∠D  respectively.

 ∠DAR + ∠ADP = 90° …… (ii)

Now, in ΔADR, by angle sum property of a triangle, we get:

∠DAR + ∠ADP + ∠AQD = 180° 

From equation (i), we get:

 90° +∠AQD = 180°

∠AQD = 90° 

We know that ∠AQD and ∠PQR are vertically opposite angles, thus,

∠PQR = 90°

Similarly, we can prove that ∠PSR = 90° .

Therefore, PQRS is a rectangle.

Hence, the correct choice is (c).