In a ΔABC, ∠CAB is an obtuse angle. P is the circumcentre of ∆ABC. Prove that ∠CAB – ∠PBC = 90°. - Geometry Mathematics 2

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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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2023-2024 (March) Model set 3 by shaalaa.com

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