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  • ABC is a right triangle with AB = AC. Bisector of angle A meets BC at D. Prove that BC = 2 AD.

ABC is a right triangle with AB = AC. Bisector of \angleA meets BC at D. Prove that BC = 2 AD.

Answers (1)

Given: ABC is a right triangle AB = AC

The Bisector of \angleA meets BC at D.

To prove: BC = 2AD

Proof : In \triangleABC, AB = AC     (given)

Then \angleB = \angleC                        (If sides are equal in a triangle, then opposite angles are also equal)

Now,

\angleBAC + \angleABC + \angleBCA = 180°                   (angle sum property)

\Rightarrow 90° + 2\angleABC = 180°                                (Q \angleABC = \angleACB)

\Rightarrow \angleB = 45° = \angleACB

\because \angleBAD = \angleCAD                                                     (AD is bisector ÐA)

BD = AD, CD = AD                                                    (Sides opposite to equal angles are equal).

Adding both,

BD + CD = AD + AD

BC = AD + AD

BC = 2AD                                                                               (\because BD + DC = BC)

Hence Proved.

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