Question:
To prove that two chords of a circle intersecting each other can not bisect(bisect equally) each other until both the chords are diameter of a circle.
Two chords AB and AC of a circle are equal. To prove bisector(bisecting equally) of angle BAC move through the centre.
Answer:
Let us suppose that Ab and AC are two equal chords of a circle.
Centre of the circle O lies on the bisector of the ∠BAC
Now join BC.
Again let the bisector of ∠BAC intersect BC at P as shown in the figure.
Now from ΔAPB and ΔAPC
AB = AC (given)
∠BAP = ∠CAP (given)
AP = AP (common)
By SAS congruent criterian
ΔAPB ≅ ΔAPC
By CPCT
BP = CP and ∠APB = ∠APC
Now ∠APB + ∠APC = 180 (since linear pairs)
=> 2∠APB = 180
=> ∠APB = 180/2
=> ∠APB = 90
So BP = CP and ∠APB = 90
Hense AP is the perpendicular bisector of the chord BC and AP passes through the centre of the circle.
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