Multiple choice

If $H$ is the orthocentre of a triangle $ABC$, then the radii of the circle circumscribing the triangles $BHC, CHA$ and $AHB$ respectively equal to:

  1. $R, R, R$
  2. $\sqrt { 2 } R,\sqrt { 2 } R,\sqrt { 2 } R$
  3. $2R,2R,2R$
  4. $\cfrac { R }{ 2 } ,\cfrac { R }{ 2 } ,\cfrac { R }{ 2 } $
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A Correct answer
Explanation

The circumradius of a triangle formed by the orthocenter and two vertices is equal to the circumradius of the original triangle ABC. This is a known geometric property.

AI explanation

If R is the circumradius of triangle ABC, the extended law of sines gives a equals 2R sin(A). Because angle BHC equals 180 degrees minus angle A, sin(BHC) equals sin(A). Applying the law of sines to triangle BHC with its circumradius r, HC divided by sin(BHC) equals 2r; substituting HC equals 2R sin(B) reveals that r equals R, and this symmetric logic applies to the other two triangles.

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