Q. 4

# AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the center of the circle.

Let AB and BC be equal chords of circle with center O.

Draw the angle bisector AD of BAC. Join BC, meeting AD at M.

In triangle DAM and CAM:

AB = BC (given that the chords are equal)

BAM = CAM ( AD is angle bisector of A)

AM = AM (common side)

ΔABD = ΔACD (by SAS congruence rule)

BM = CM and AMB = AMC = x (say) (by CPCT) ………(1)

But AMB + AMC = 180° ………(2)

From equation (1) and (2), we have:

x + x = 180°

2x = 1280°

x = 90°

AMB = AMC = 90°

Therefore, AD is the perpendicular bisector of chord BC, but the perpendicular bisector of any chord always passes from the center of the circle.

AD passes through the center O of the circle.

Thus, O lies on the angle bisector of the angle BAC.

Hence, proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  RD Sharma | Extra Qs. of Cyclic Quadrilaterals31 mins  Important Questions on Circles29 mins  NCERT | Imp. Qs. on Circles43 mins  Quiz | Lets Roll on a Circle45 mins  Proof of Important Theorems of Circles45 mins  Concept of Cyclic Quadrilateral61 mins  Master Theorems in Circles42 mins  Genius Quiz | Rolling Around Circles49 mins  Proof of Important Theorems of Circles46 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 