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Question
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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