Question

In a parallelogram ABCD; BX is the bisector of ∠ABC and DY is the bisector of ∠ADC. Prove that BXDY is a parallelogram.

Answer 1

Given: ABCD is a parallelogram. BX is the bisector of ∠ABC meeting AD at X. DY is the bisector of ∠ADC meeting BC at Y.

To Prove: Quadrilateral BXDY is a parallelogram.

Proof [Step-wise]:

1. In parallelogram ABCD, opposite sides are parallel:

AB || CD and AD || BC

2. Opposite angles of a parallelogram are equal.

So, ∠ABC = ∠ADC.

3. BX bisects ∠ABC,

So, `∠ABX = 1/2 ∠ABC`.

DY bisects ∠ADC,

So, `∠CDY = 1/2 ∠ADC`. 

From step 2,

`1/2 ∠ABC = 1/2 ∠ADC` 

Hence, ∠ABX = ∠CDY.

4. AB || CD (step 1).

Because BX makes the same angle with AB that DY makes with CD and AB || CD, it follows that BX || DY if two lines make equal angles with two parallel lines, the two lines are parallel.

5. By construction X lies on AD, so XD is collinear with AD.

Y lies on BC, so YB is collinear with BC.

Since AD || BC (step 1), we have XD || BY.

6. From steps 4 and 5, we have both pairs of opposite sides of quadrilateral BXDY parallel BX || DY and XD || BY. 

Therefore, BXDY is a parallelogram.