Question

If $R$ be circum radius of $\Delta ABC,$ then circum radius of it's pedal triangle is:

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

Answer 1

The correct option is D $\frac{R}{2}$

In $\Delta BFC$
$BF=a\cos B$
Also point $B,F,E,A$ are concyclic.
$\angle FBE=\angle ABE={90}^0-A$
Similarly point $B,F,E,C$ are concyclic.
$\angle BEF=\angle BCF={90}^0-B$
Then in $\Delta BEF$
By sine rule,
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\cdots \left(i\right)$
$\frac{EF}{\sin \left(90-A\right)}=\frac{a\cos B}{\sin \left(90-B\right)}$
$\frac{EF}{\cos A}=\frac{a\cos B}{\cos B}$
$EF=a\cos A$
Similarly,
$DE=c\cos C$ and $FD=b\cos B$
For circum-radius $R^{\prime }$ of $\Delta DEF$
$2R^{\prime }=\frac{EF}{\sin \angle EDF}$
$2R^{\prime }=\frac{a\cos A}{\sin ({180}^0-2A)}$
$2R^{\prime }=\frac{a\cos A}{\sin 2A}$
$2R^{\prime }=\frac{a\cos A}{2\sin A\cos A}$
$2R^{\prime }=\frac{a}{2\sin A}$
$2R^{\prime }=\frac{2R}{2}[$From equation $\left(i\right)]$
$R^{\prime }=\frac{R}{2}$