Let (O) and (H) be the circumcenter and orthocenter of an acute angled triangle (ABC). Prove that the area of one of the triangles (AOH, BOH) and (COH) is the sum of the areas of the other two.

This topic was modified 3 months, 3 weeks ago by swastik pramanik.

Now if not all three of them are 0, then there are two of the above three signed area have same sign. WLOG let’s consider (BOH) and (COH) have same sign, then

$$|AOH|=|-(BOH)-(COH)|=|BOH+COH|=|BOH|+|COH|$$

Hence we have our desired result.

This reply was modified 3 months, 3 weeks ago by Nitin Prasad.

This reply was modified 3 months, 3 weeks ago by Nitin Prasad.

This reply was modified 3 months, 3 weeks ago by Nitin Prasad.