In the figure, AB = AC, CH = CB and HK || BC. If ∠CAX = 137°, find ∠CHK.

Given: AB = AC, CH = CB, HK || BC and CAX = 137°

In the above figure, ∠CAX and ∠CAH form a linear pair.

⇒ ∠CAX + ∠CAH = 180°

⇒ ∠CAH = 180° – ∠CAX

⇒ ∠CAH = 180° – 137°

∴ ∠CAH = 43°

Now, consider ΔABC.

We know, the sum of angles in a triangle is 180°.

⇒ ∠ABC + ∠ACB + ∠CAB = 180°

But given AB = AC ⇒ ∠ABC = ∠ACB

We have ∠CAB = 43° and let ∠ABC = ∠ACB = x

⇒ x + x + 43° = 180°

⇒ 2x = 137°

∴ ∠ABC = ∠ACB = 68.5°

Now, consider ΔHBC.

Given, CH = CB ⇒ ∠CHB = ∠HBC

But, ∠HBC = ∠ABC (same angle)

∴ ∠CHB = 68.5°

We again use the sum of angles in a triangle is 180°.

⇒ ∠CHB + ∠HBC + ∠BCH = 180°

⇒ 68.5° + 68.5° + ∠BCH = 180°

⇒ ∠BCH = 180° – 137°

∴ ∠BCH = 43°

But, we also have HK || BC.

⇒ ∠CHK = ∠BCH (alternate interior angles)

∴ ∠CHK = 43°

Thus, ∠CHK = 43°