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If AD, BE, CF are the diameters of circumcircle of Delta ABC, then prove that area of hexagon AFBDCE is 2Delta |
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Answer» Solution :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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