Question

In the adjoining figure, ABCD is a parallelogram. AX and CY are the bisectors of ∠A and ∠C respectively. 

Prove that :

1. ΔADX ≅ ΔCBY
2. AX || CY
3. AYCX is a parallelogram.

Answer 1

Given:

ABCD is a parallelogram.

AX is the bisector of ∠A.

CY is the bisector of ∠C.

From ABCD being a parallelogram, we may use:

AD || BC

AB || CD

AD = BC

AB = CD

And ∠A = ∠C, ∠B = ∠D

To Prove:

1. △ADX ≅ △CBY 
2. AX || CY 
3. AYCX is a parallelogram

Proof [Step-wise]:

1. Parallelogram facts:

AD = BC   ...(Opposite sides of a parallelogram)

AB || CD and AD || BC

∠A = ∠C   ...(Opposite angles of a parallelogram)

2. Angle relations from bisectors:

AX bisects ∠A

⇒ ∠DAX = ∠XAB

⇒ `∠DAX = 1/2 ∠A`

CY bisects ∠C

⇒ ∠BCY = ∠YCD

⇒ `∠BCY = 1/2 ∠C`

Since ∠A = ∠C, we get

∠DAX = ∠BCY

Thus, one pair of corresponding angles in the two triangles are equal.

3. Angles at D and B:

X lies on DC

So, ∠ADX = ∠ADC angle formed by AD and DX where DX is along DC.

Y lies on AB

So, ∠CBY = ∠CBA angle formed by BC and BY where BY is along BA.

In a parallelogram ∠ADC = ∠CBA since ∠D = ∠B.

Hence, ∠ADX = ∠CBY.

4. Establish congruence (ASA):

In △ADX and △CBY:

1. AD = BC   ...(Step 1) 
2. ∠ADX = ∠CBY   ...(Step 3) 
3. ∠DAX = ∠BCY   ...(Step 2)

Therefore, by ASA, △ADX ≅ △CBY.

This proves (i).

5. From congruence / angle-equality to parallelism:

From step 2, we had ∠DAX = ∠BCY.

But AD || BC (Step 1).

So, if the angle that AX makes with AD equals the angle that CY makes with BC, the lines AX and CY are parallel, corresponding/alternate interior angle reasoning.

Hence, AX || CY.

This proves (ii).

6. Prove AYCX is a parallelogram:

We already have AX || CY (Step 5).

AY is a part of AB and CX is a part of CD.

Since AB || CD (Step 1), AY || CX.

Both pairs of opposite sides of quadrilateral AYCX are parallel.

So, AYCX is a parallelogram. 

This proves (iii).