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).

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Hope it will be helpful :)

Answer 2

Solution

The region between a chord and either of its arcs is called a segment the circle.

Angles in the same segment of a circle are equal.

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Let bisectors of ∠A  meet the circumcircle of ∆ABC at M.

Join BM & CM

∠MBC = ∠MAC

[Angle in the same segment are equal]

 

∠BCM = ∠BAM

[Angle in the same segment are equal]

But , ∠BAM= ∠CAM..........(i)

[Since, AM is bisector of  ∠A]

∠MBC= ∠BCM

MB= MC

[SIDE OPPOSITE TO EQUAL ANGLES ARE EQUAL]

So,M must lie on the perpendicular bisector of BC.

Let, M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ABC. Join AM.

 

Since, M lies on perpendicular bisector of BC.

BM= CM

∠MBC= ∠MCB

But ∠MBC= ∠MAC

[Angles in same segment are equal]

∠MCB= ∠BAM

[Angles in same segment are equal]

From eq i

∠BAM= ∠CAM

So, AM is the bisector of ∠A

Hence, bisector of ∠A and perpendicular bisector BC intersect at M which lies on the circumcircle of the ∆ABC..

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Hope it helps uhh.