Question

Que - .In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.​

Answer 1

Answer:

Solution:

Here, join BE and CE.

Now, since AE is the bisector of ∠BAC,

• ∠BAE = ∠CAE

Also,

∴ arc BE = arc EC

This implies chord BE = chord EC

Now, for triangles ΔBDE and ΔCDE,

• DE = DE (It is the common side)

• BD = CD (It is given in the question)

• BE = CE (Already proved)

So, by SSS congruency, ΔBDE ≌ ΔCDE.

Thus, ∴∠BDE = ∠CDE

We know, ∠BDE = ∠CDE = 180°

Or, ∠BDE = ∠CDE = 90°

∴ DE ⊥BC (hence proved).

Hope it will be helpful :)

Answer 2

Answer:

Suppose angle bisector of angle A meets circumcircle with centre O at P, and OP is joined to meet BC at D.                                                            ...(i)

 Angle BAP = Angle CAP (Because AP is angle bisector)

Angle BOP = 2 Angle BAP

Angle COP = 2 Angle CAP

So, Angle BOP = Angle COP

Therefore in triangles BDO and CDO,

BO = CO                          (Radii of the same circle)

Angle BOP = Angle COP (Proven above)

DO = DO                          (Common side

So triangles BDO and CDO are congruent. (SAS criterion)

So, BD = CD and angle BDO = angle CDO (CPCTE)

But angles BDO and CDO form linear pair.

So, angle BDO + angle CDO = 180 degree

So, angle BDO = half of 180 = 90 degree

With BD = CD and angle BDO = 90 degree,

OD is perpendicular bisector of BC

OR

OP is perpendicular bisector of CD                                                  ...(ii)

By (i) and (ii),

P lies on angle bisector of angle A and perpendicular bisector of BC and is a point

Hence proven.

I hope is useful for u......

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