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

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#### Solution

**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.

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