Mathematics
In the adjoining figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its center O, prove that AD bisects ∠BAC and ∠BDC.
Circles
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Answer
As chords AB and AC are equidistant from the center, so AB = AC. [∵ equal chords are equidistant from center]
Since, angle in a semicircle = 90°.
∠B = ∠C
In △ABD and △ACD,
∠B = ∠C (Both equal to 90°)
AD = AD (Common)
AB = AC (Proved above)
∴ △ABD ≅ △ACD (By R.H.S. congruence rule).
∴ ∠BAD = ∠CAD and ∠BDA = ∠CDA (By C.P.C.T.)
Hence, proved that AD bisects ∠BAC and ∠BDC.
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