Question

In triangle ABC, angle A is an obtuse angle.p is the circumcentre of triangles ABC of triangles ABC . Provethat angle PBC = angleA-90°

Answer 1

The image of the problem is given below.

In ΔABC,

∠A is an obtuse angle.

Radii of the same circle are equal.

⇒ PA = PB = PC

In ΔPBC,      PB = PC

In isosceles triangle, two angles are equal.

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

In ΔPAB,      PA = PB

In isosceles triangle, two angles are equal.

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

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

In ΔPAC,      PA = 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 (1) and (2),

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

$\text { since } \angle \mathrm{PBC}=\angle \mathrm{PCB}$

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

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

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

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

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

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

Hence proved.

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