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If AD, BE, CF are the diameters of circu...

If AD, BE, CF are the diameters of circumcircle of `Delta ABC`, then prove that area of hexagon AFBDCE is `2Delta`

Text Solution

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Chord AC subtends the same angle at point B and D
`:. Angle ADC = B`
So, in right angled triangled ACD
`CD = AC cot B`
`:. CD = b cot B`
Similarly from `Delta ABD`,
`BD = c cot C`

`:. " Area of " Delta BDC`
`Delta_(BDC) = (1)/(2) (b cot B) (c cot C) sin (B + C)`
`= (bc)/(2) sin A cot C cot B`
`= (abc)/(4R) cot B cot C`
Similarly `Delta_(AEC) = (abc)/(4R) cot A cot C`
and `Delta_(AFB) = (abc)/(4R) cot A cot B`
`:.` Area of hexagon AFBDCE
`= Delta_(BDC) + Delta_(AFC) + Delta_(AFB) + Delta_(ABC)`
`= (abc)/(4R) [cot A cot B + cot B cot C + cot C cot A] + Delta`
`= Delta + Delta`
(as in `Delta ABC, cot A cot B + cot B cot C + cot A = 1`)
`=2 Delta`
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