Question

In a ΔABC, ∠CAB is an obtuse angle. P is the circumcentre of ∆ABC. Prove that ∠CAB – ∠PBC = 90°.

Answer 1

Given: ∠CAB is an obtuse angle and P is the circumcentre of ΔABC.

Construction: Draw BD as diameter, join AD.

Proof: ∠CAD = ∠CBD   ......[Angles on same arc]

⇒ ∠CAD = ∠CBP  ......(i)

Also, ∠BAD = 90°  ......(ii) [Angle in semi-circle]

Now, from figure,

∠CAB = ∠CAD + ∠DAB

⇒ ∠CAB = ∠CBP + 90°  ......[Using (i) and (ii)]

⇒ ∠CAB – ∠CBP = 90° 

or ∠CAB – ∠PBC = 90°.

Hence proved.