Question

In a trapezium ABCD, the diagonals AC and BD intersect at point ‘O’, AB || DC and AD = BC.

Prove that :

1. ∠BAD = ∠ABC
2. ∠ADC = ∠BCD
3. AC = BD
4. OA = OB and OC = OD

Answer 1

Given:

ABCD is a trapezium with AB || DC.

AD = BC.

The diagonals AC and BD meet at O.

To Prove:

1. ∠BAD = ∠ABC 
2. ∠ADC = ∠BCD 
3. AC = BD 
4. OA = OB and OC = OD

Proof [Step-wise]:

1. Show ∠BAD = ∠ABC.

Construction/idea: Produce AB to a point E and draw CE || AD so AE || DC. 

Then ADCE is a parallelogram, so CE = AD.

Because CE = AD and BC = AD (given), we get CE = BC.

Hence, triangle CEB is isosceles and ∠EBC = ∠BEC. 

Using the parallelism relations and angle sums one obtains ∠A = ∠B.

i.e., ∠BAD = ∠ABC.

2. Show ∠ADC = ∠BCD.

From (1) we have ∠A = ∠B.

In any quadrilateral, the sums of opposite angles are supplementary.

Using AB || DC co-interior angles, we get 

∠A + ∠D = 180° and ∠B + ∠C = 180°

Since ∠A = ∠B. 

Subtracting gives ∠D = ∠C.

i.e., ∠ADC = ∠BCD.

3. Show AC = BD.

From (1), we have ∠A = ∠B.

In triangles ABC and BAD:

AD = BC   ...(Given)

∠A = ∠B   ...(Proved)

AB = AB   ...(Common)

So, ΔABC ≅ ΔBAD by ASA.

Hence, corresponding parts give AC = BD. 

4. Show OA = OB and OC = OD.

Consider the intersection O of diagonals AC and BD.

Because AB || DC, triangles AOB and COD are similar AA:

Alternate interior angles and vertical angles.

Hence, `(AO)/(CO) = (BO)/(DO) = (AB)/(CD)`. 

Also, with the same parallelism, triangles AOD and COB are similar AA. 

From that similarity, we get

`(AO)/(BO) = (AD)/(BC)` and `(DO)/(CO) = (AD)/(BC)`

Since AD = BC   ...(Given)

`(AO)/(BO) = 1` and `(DO)/(CO) = 1`

Therefore, AO = BO and DO = CO.

i.e., OA = OB and OC = OD.