MathLabs

Problem 2

Let OO be the circumcentre and HH the orthocentre of an acute triangle ABCABC. Prove that the area of one of △AOH\triangle AOH, △BOH\triangle BOH, and △COH\triangle COH equals the sum of the areas of the other two.
Step 4 of 4: Compare the three areas
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[BOH]+[COH]=OH⋅2d(M,OH)2=OH⋅d(A,OH)2=[AOH][BOH]+[COH]=\frac{OH\cdot2d(M,OH)}2=\frac{OH\cdot d(A,OH)}2=[AOH]
Detailed analysis

Using the base OHOH, [AOH]=OH⋅d(A,OH)/2[AOH]=OH\cdot d(A,OH)/2. Thus [BOH]+[COH]=OH⋅d(M,OH)=OH⋅d(A,OH)/2=[AOH][BOH]+[COH]=OH\cdot d(M,OH)=OH\cdot d(A,OH)/2=[AOH]. Therefore one area equals the sum of the other two.