Question

In the circle shown alongside,the chords AB and AC are of same length. The bisector of $\angle$A intersects the chord BC at D and meets the circle at E. Then which of the following is/are true?

• A. D is the midpoint of BC.
• B. D divides BC in the ratio 1:2.
• C. AE is perpendicular bisector of chord BC
• D. AE is a diameter

Answer 1

Answer: (A), (C), (D)

The correct options are
A

D is the midpoint of BC.

C

AE is perpendicular bisector of chord BC

D

AE is a diameter

In $△$ACD and $△$ABD,

$\angle$CAD = $\angle$BAD (angle bisector)

$\angle$C = $\angle$B (as AB = AC)

AC = AB (given)

$\therefore$ $△$ACD $\cong$ $△$ABD (by ASA)

$⟹$ CD = BD (cpct)

Therefore D is the midpoint of BC.

Also, $\angle$ADC = $\angle$ADB = $x$(say) (cpct)

$x+x={180}^{\circ }$ (linear pair)

$⟹x={90}^{\circ }$

Therefore AE is perpendicular bisector of chord BC.

Hence AE must pass through the centre of circle.

AE is a diameter.