Problem 2

Let

O

be the circumcentre and

H

the orthocentre of an acute triangle

ABC

. Prove that the area of one of

\triangle AOH

\triangle BOH

, and

\triangle COH

equals the sum of the areas of the other two.

Step 2 of 4: Average the distances at the midpoint

Previous step Next step

M\text{ midpoint of }BC,\qquad d(M,OH)=\frac{d(B,OH)+d(C,OH)}2

Detailed analysis

Let $M$ be the midpoint of $BC$ . Since distance to a fixed line is affine along a segment whose endpoints lie on the same side, $d(M,OH)=(d(B,OH)+d(C,OH))/2$ . Therefore $[BOH]+[COH]=OH\cdot d(M,OH)$ .