Math, asked by vjpawar1977, 1 year ago

In triangle ABC measure of angle CAB is an obtuse. P is the circumcentre of triangle ABC. Prove that angle PBC=angle CAB -90​

Answers

Answered by presentmoment
4

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}, \quad \mathrm{PB}=\mathrm{PC}

Two angles are equal in isosceles triangle.

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

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

Two angles are equal in isosceles triangle.

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

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

\text { In } \triangle \mathrm{PAC}, \quad \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)

Add (1) and (2), we get

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

Since \angle \mathrm{PBC}=\angle \mathrm{PCB}

\begin{array}{l}{\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}}\end{array}

Hence proved.

To learn more...

brainly.in/question/15174796

brainly.in/question/14988261

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Answered by darinjoy4105
1

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