Question

Angle CAB of Triangle ABC is obtuse angle. P is circumcentre
of Triangle ABC.prove that angle PBC=angleCAB-90

Answer 1

Let the circumcentre of △ ABC = P

∠A is an obtuse angle

Therefore,  PA = PB = PC ( radii of same circle)

In ΔPBC,

PB = PC ( radii of same circle)

∠PBC = ∠PCB (Angles opposite to the equal sides are equal)

In ΔPAB,

PA = PB  ( radii of same circle)

∠PAB = ∠PBA (Angles opposite to the equal sides are equal)

∠PAB = ∠PBC + ∠ ABC  --- eq 1

In ΔPAC,

PA = PC  ( radii of same circle)

∠PAC =  PCA (Angles opposite to the equal sides are equal)

∠ PAC = ∠ PCB + ∠ ACB  --- eq 2

Adding equation (1) and (2),

= ∠ PAB + ∠ PAC =  ∠ PBC + ∠ ABC + ∠ PCB + ∠ ACB

= ∠ BAC = ∠PBC + (∠ ABC + ∠ ACB) + ∠ PBC    [As ∠ PBC = ∠ PCB]

= ∠ BAC = 2∠PBC + (180° − ∠ BAC)

= 2∠BAC = 2∠PBC + 180°

= ∠BAC = ∠PBC + 90°

= ∠PBC = ∠BAC − 90°

= ∠PBC = ∠A − 90°                      

Hence proved