can you please prove this with written proof...
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We have, <ABC=<ADC (since both are subtended by same arc AC)
Then, <AOC= 2<ABC (Since, angle subtended by an arc on the centre is double the angle subtended by it any point on the circle)
=<AOC= <ABC + <ADC -(i)
But, <AFC= <ABC +<BAD ( Since, ext. angle= sum of two interior angles)
Therefore, <AOC= <AFC - <BAD + <ADC -(ii)
but, <ADC - <BAD =<AEC -(iii)
From, (ii) and (iii), we get
<AOC= <AFC +<AEC
Then, <AOC= 2<ABC (Since, angle subtended by an arc on the centre is double the angle subtended by it any point on the circle)
=<AOC= <ABC + <ADC -(i)
But, <AFC= <ABC +<BAD ( Since, ext. angle= sum of two interior angles)
Therefore, <AOC= <AFC - <BAD + <ADC -(ii)
but, <ADC - <BAD =<AEC -(iii)
From, (ii) and (iii), we get
<AOC= <AFC +<AEC
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