Question

In triangle ABC angle CAB is an obtuse angle P is the circumcentre of triangle ABC then prove that angle PBC=angle CAB-90°

Answer 1

The image of the problem is given below.

$\text { In } \triangle \mathrm{ABC},$

∠CAB is an obtuse angle.

Let P be the circumcentre of the triangle ABC.

Radii of the same circle are equal.

⇒ PA = PB = PC

$\text { In } \triangle \mathrm{PBC}, \ \ \ \mathrm{PB}=\mathrm{PC}$

Angle opposite to equal sides are equal.

$\Rightarrow \angle \mathrm{PBC}=\angle \mathrm{PCB}$

$\mathrm{In} \triangle \mathrm{PAB}, \ \ \ \mathrm{PA}=\mathrm{PB}$

Angle opposite to equal sides are equal.

$\Rightarrow \angle \mathrm{PAB}=\angle \mathrm{PBA}$

$\Rightarrow \angle \mathrm{PAB}=\angle \mathrm{PBC}+\angle \mathrm{ABC}$ – – – – (1)

$\text { In } \triangle \mathrm{PAC}, \ \ \ \mathrm{PA}=\mathrm{PC}$

Angle opposite to equal sides are equal.

$\Rightarrow \angle \mathrm{PAC}=\angle \mathrm{PCA}$

$\Rightarrow \angle \mathrm{PAC}=\angle \mathrm{PCB}+\angle \mathrm{ACB}$ – – – – (2)

Adding equation (1) and equation (2), we get

$\angle \mathrm{PAB}+\angle \mathrm{PAC}=\angle \mathrm{PBC}+\angle \mathrm{ABC}+\angle \mathrm{PCB}+\angle \mathrm{ACB}$

We know that $\angle \mathrm{PBC}=\angle \mathrm{PCB}$

$\angle \mathrm{BAC}=\angle \mathrm{PBC}+(\angle \mathrm{ABC}+\angle \mathrm{ACB})+\angle \mathrm{PBC}$

$\Rightarrow \angle \mathrm{BAC}=2 \angle \mathrm{PBC}+\left(180^{\circ}-\angle \mathrm{BAC}\right)$

$\Rightarrow 2 \angle \mathrm{BAC}=2 \angle \mathrm{PBC}+180^{\circ}$

$\Rightarrow \angle B A C=\angle P B C+90^{\circ}$

$\Rightarrow \angle \mathrm{PBC}=\angle \mathrm{BAC}-90^{\circ}$

$\Rightarrow \angle \mathrm{PBC}=\angle \mathrm{CAB}-90^{\circ}$

Hence proved.

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Answer 2

Given : ∠CAB is an obtuse angle

To prove : ∠CAB is an obtuse angle

Solution:

in triangle ABD

∠CAB is an obtuse angle

PA=PB=PC (radii)

in triangle PBC

PB = PC

(angle opp to equal side are equal )

∠PBC = ∠PBA

∠PAB = ∠PBC + ∠ABC ..(1)

similarly

∠PAC = ∠PCB +∠ACB ..(2)

add 1 and 2 we get

∠PAB  +∠PAC = ∠PBC + ∠ABC+∠PCB +∠ACB

as

∠PBC = ∠PCB

∠BAC = ∠PBC +(∠ABC+∠ACB)+∠PBC

∠BAC = 2∠PBC + (180° - ∠BAC)

2∠BAC =2∠PBC + 180°

∠PBC = ∠CAB -90°

HENCE PROVED

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