AB and AC are two equal chords of circle .prove that bisector of angle BAC pass through centre
Answers
Answer: Proved
Step-by-step explanation:
Let the two chords AB and AC are chords
centre of the circle o lies on the bisector of the ∠BAC
Now join BC
let the bisector of ∠ B A C intersect BC at P
Now from ΔAPB and ΔAPC
AB =AC
BAP =CAP
AP=AP
ΔAPB ≅ Δ APC
BP=CP and ∠APB=∠ APC ( By CPCT)
Now ∠APB + ∠APC = 180° ( linear pair)
2 ∠APB = 180
∠APB = 180 / 2
∠APB = 90°
So. BP =CP ,
Hence AP is the perpendicular bisector of the chord BC and AP passes through the centre of the circle .
Answer:
Step-by-step explanation:
Let the two chords AB and AC are chords
centre of the circle o lies on the bisector of the ∠BAC
Now join BC
let the bisector of ∠ B A C intersect BC at P
Now from ΔAPB and ΔAPC
AB =AC
BAP =CAP
AP=AP
ΔAPB ≅ Δ APC
BP=CP and ∠APB=∠ APC ( By CPCT)
Now ∠APB + ∠APC = 180° ( linear pair)
2 ∠APB = 180
∠APB = 180 / 2
∠APB = 90°
So. BP =CP ,
Hence AP is the perpendicular bisector of the chord BC and AP passes through the centre of the circle .