Question

In the adjoining figure, AB = AC, AP = AQ. Prove that :

1. ΔAPC ≅ ΔAQB
2. CP = BQ
3. ∠APC ≅ ∠AQB

Answer 1

Given:

• In triangle ABC, AB = AC.
• Points P and Q lie on AB and AC respectively, with AP = AQ.

To Prove:

1. ΔAPC ≅ ΔAQB.
2. CP = BQ.
3. ∠APC ≅ ∠AQB.

Proof [Step-wise]:

1. AP = AQ   ...(Given)

2. AB = AC   ...(Given)

3. P lies on AB and Q lies on AC, so AP is collinear with AB and AQ is collinear with AC.

Therefore, the angle between AP and AC equals the angle between AB and AC.

i.e. ∠PAC = ∠QAB = ∠BAC   ...(Collinearity/definition of the angles)

4. In ΔAPC and ΔAQB we have:

AP = AQ   ...(Step 1)

AC = AB   ...(Step 2)

∠PAC = ∠QAB   ...(Step 3)

Thus, the two triangles have two sides and the included angle is equal.    ...(SAS criterion)

5. Therefore, ΔAPC ≅ ΔAQB by SAS.   ...(From step 4)

6. From the congruence (CPCTC - corresponding parts of congruent triangles are equal):

CP = BQ   ...(Corresponding sides)

∠APC = ∠AQB   ...(Corresponding angles)