# Radii of the circles circumscribing the triangles $BHC,CHA,AHB,ABC$ are all equal

If $H$ is the orthocenter of a triangle $ABC$;prove that the radii of the circles circumscribing the triangles $BHC,CHA,AHB,ABC$ are all equal.

• Are you familiar with the extended law of sines? $$2R=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$ – Sawarnik Aug 7 '15 at 16:03

## 2 Answers

Since $\widehat{AHB}+\widehat{ACB}=\pi$, $AHB$ and $ABC$ have the same circumradius by the sine theorem, since they have $AB$ in common and $\sin\widehat{AHB}=\sin\widehat{ACB}$.

Hint: Let $H_A$, $H_B$, and $H_C$ be the reflections of $H$ about $BC$, $CA$, and $AB$, respectively. Prove that $H_A$, $H_B$, and $H_C$ are on the circumscribed circle of $ABC$.

• How will i prove that?I am confused,can you elaborate me,please? – Brahmagupta Aug 7 '15 at 15:23
• What is the angle $\angle BH_AC=\angle BHC$, for instance? The answer should be in terms of $\angle BAC$, $\angle CBA$, and $\angle ACB$. – Batominovski Aug 7 '15 at 15:25
• Why are they equal,Sir?Is this a property of reflection in a line.And if they are equal,what how are they on the circumcircle of ABC?Sorry to bother you. – Brahmagupta Aug 7 '15 at 15:29
• A reflection is an isometry, and every isometry preserves angles. In this particular case, $B$ and $C$ are fixed points of the reflection about $BC$. What is the image of the triangle $BHC$ under the reflection about $BC$? Finally, I asked you to calculate the angle $\angle BHC$ to see why $\angle BH_AC$ is on the circumcircle of $ABC$. – Batominovski Aug 7 '15 at 15:35
• – Blue Aug 7 '15 at 15:51