Question

In a $\Delta ABC$, if $AD,BE$ and $CF$ are the altitudes and R is the circum-radius, then the radius of the circumcentre $DEF$ is?

• A. $\frac{R}{2}$
• B. $2R$
• C. $R$
• D. $\frac{3}{2}R$

Answer 1

Answer: (A)

$\frac{R}{2}$

$BE$ and $CF$ are altitudes of triangle $ABC$ .Therefore the circle described on $BC$ as diameter passes through $E$ and $F$ and have radius $\frac{a}{2}$

$\angle FBC={90}^{\circ }-A$

So by using sine rule

$FE=2\left(\frac{a}{2}\right)\sin {(90}^{\circ }-A)FE=a\cos A$

Let the radius of the circumcircle of $△DEF$ be $x$

$x=\frac{FE}{2\sin \angle FDE}x=\frac{a\cos A}{2\sin ({180}^{\circ }-2A)}x=\frac{a\cos A}{2\sin 2A}=\frac{a}{4\sin A}=\frac{R}{2}$