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D, E, F are three points on the sides BC, CA, AB, respectively, such that angleADB = angle BEC = angle CFA= theta. A', B', C' are the points of intersections of the lines AD, BE, CF inside the triangle. Show that area of Delta A'B'C' = 4 Delta cos^(2) theta, where Delta is the area of Delta ABC |
Answer» SOLUTION :From `Delta AB'C (AB')/(sin(pi-(A + theta))) = (AC)/(sin(pi - B))`  `rArr AB' = 2R sin (A + theta)` From `Delta AC'B, (AC')/(sin (theta - A)) = (AB)/(sin (pi - C))` `rArr AC' = 2R sin (theta - A)` `:. B'C' = 2R (sin (A + theta) - sin (theta - A))` `= 4R cos theta sin A = 2a cos theta` Similarly, `C'A' = 2b cos theta` `:. " Area of " Delta A'B'C' = (1)/(2) (B'C') (A'C') sin angleB'C'A'` `= (1)/(2) (2 a cos theta) (2b cos theta) sin C` `= 4 cos^(2) theta XX Delta` |
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