Saved Bookmarks
| 1. |
Let O be the circumcentre and H be the orthocentre of an acute angled triangle ABC. If A gt B gt C, then show that Ar (Delta BOH) = Ar (Delta AOH) + Ar (Delta COH) |
Answer» SOLUTION : From the figure, we have `ANGLE OAH = A - 2(90^(@) - B) = B - C` Similarly, `angle OBH = A - C` and `angle OCH = A - B` Also, `AH = 2R cos A, BH = 2R cos B, CH = 2R cos C` `Ar (Delta AOH) = (1)/(2) (R) (2R cos A) sin (B - C)` `= R^(2) cos (B + C) sin (C - B)` `=(R^(2))/(2) (sin 2 C - sin 2B)` Similarly, `Ar(DeltaBOH) = (R^(2))/(2) (sin 2C - sin 2A)` and `Ar (Delta COH) = (R^(2))/(2) (sin 2 B - sin 2A)` Clearly, `AR (Delta AOH) + Ar (Delta COH) = Ar (Delta BOH)` |
|