Question

In the adjoining figure, AB = CD and ∠ABC = ∠BCD. Prove that : 

1. AC = BD 
2. BP = CP.

Answer 1

Given: AB = CD, ∠ABC = ∠BCD and AC and BD meet at P as shown.

To Prove:

1. AC = BD 
2. BP = CP

Proof [Step-wise]:

1. Consider triangles △ABC and △DCB.

2. AB = CD.   ...(Given)

3. BC = CB.   ...(Common side)

4. ∠ABC = ∠BCD (given), this is the included angle between AB and BC in △ABC and between DC and CB in △DCB.

5. Therefore, by SAS, △ABC ≅ △DCB.

6. From the congruence (CPCTC), the corresponding sides AC and BD are equal.

Hence, AC = BD. 

This proves (i).

7. From the congruence we also get ∠BCA = ∠DBC corresponding angles.

8. Note that ∠PBC is the angle between PB (part of BD) and BC, so ∠PBC = ∠DBC.

Likewise, ∠BCP is the angle between BC and PC (part of AC), so ∠BCP = ∠BCA.

9. Using step 7, ∠PBC = ∠BCP.

Thus, in triangle PBC the base angles at B and C are equal.

So the sides opposite them are equal: BP = CP.

This proves (ii).

(i) AC = BD and (ii) BP = CP, as required.