Question

please give me answer

Answer 1

Heyy Buddy❤

Here's your Answer...

PROOF;

IN TRIANGLE BEC,

ABC = BEC + BCE ( Angle sum property of Exterior angle)

=> Squaring on both sides.

=> 2ABC = 2 ( BEC + BCE) -------(1).

Now,

We know that angle substended at the circumference is half of the angle substended at the centre..

=>2 ABC = AOC --------(2)

From eq 1 and 2, we get,

AOC = 2 ( BEC + BCE)

=> AOC = BEC + BEC + BCE + BCE

=> AOC = BCE + BEC + ABC ( ABC = BEC + BCE)

=> AND, ABC = ADC ( Angle substended at the same arc are equal)

=> AOC = BCE + BEC +ADC ------(3)

=> In Triangle FDC

AFC = FDC + DCF

=> AFC = BCD ( BCE) + ADC. ------(4)

NOW,

putting eq 3 in eq 4.

AOC= AEC + AFC.
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