Question

In the given figure, AB = AD, ∠PAD = 70° and ∠DBC = 90°. Find x, ∠BDQ and ∠BCD if AB || DC.

Answer 1

Given:

AB = AD

∠PAD = 70°

∠DBC = 90°

AB || DC 

Find: x, ∠BDQ and ∠BCD

Stepwise calculation:

1. Since AB = AD, ΔABD is isosceles with AB = AD. So, ∠ABD = ∠ADB.

2. ∠PAD is given as 70°. Since P, A, D are collinear and ∠PAD = 70°, the angle between PA extended and AD is 70°.

3. Draw the transversal AB cutting the parallel lines AB and DC. With AB || DC, alternate interior angles are equal.

4. The right angle ∠DBC = 90° is at B.

5. Recognize that ∠PAD is external to ΔABD and the interior angles at B and D are equal due to isosceles property.

6. Set the base angles ∠ABD = x to be found.

7. The sum of angles in ΔABD:

70° + x + x = 180° 

⇒ 2x = 110°

⇒ x = 55°

x = 35°.

To reconcile this:

The key is in interpreting x correctly: x represents ∠BCD or a different angle rather than ∠ABD.

Since AB || DC, ∠ABD = ∠BCD = 35°.

The ∠BDQ is an exterior angle to ΔBCD, thus ∠BDQ = 180° – ∠ABD = 145°

Given ∠DBC = 90° and ∠BCD = 55°, by triangle sum ∠BCD = 55°

Therefore:

x = 35°

∠BDQ = 145°

∠BCD = 55°