Question

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:

• A. $R,R,R$
• B. $\sqrt{2}R,\sqrt{2}R,\sqrt{2}R$
• C. $2R,2R,2R$
• D. $\frac{R}{2},\frac{R}{2},\frac{R}{2}$

Answer 1

The correct option is A $R,R,R$

$\angle BAE=90-B$

$\angle ABF=90-A$

$\therefore \angle AHB=A+B$

Similarly,

$\angle BHC=B+Cand\angle AHC=A+C$

Let R be the circumference of $△ABC$ and $R1,R2$

and $R3$ be radil of the circumcircles of$△AHB,△AHC$ respectively

$In△AHB,$

$\angle AHB=A+B$

$\therefore R1=\frac{AB}{2\sin (A+B)}=\frac{C}{2\sin C}=R$

Similarly, $R2=R\&R3=R$

$\therefore R,R,R$.