Question

In a triangle $ABC$, points $D$ and $E$ are taken on side $BC$ such that $BD=DE=EC$. If $\angle ADE=\angle AED=\theta$, then

• A. $\tan \theta =3\tan B$
• B. $\tan \theta =3\tan C$
• C. $\tan A=\frac{6\tan \theta }{{\tan }^2\theta -9}$
• D. $9{\cot }^2\left(\frac{A}{2}\right)={\tan }^2\theta$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\tan \theta =3\tan B$
B $\tan \theta =3\tan C$
C $\tan A=\frac{6\tan \theta }{{\tan }^2\theta -9}$
D $9{\cot }^2\left(\frac{A}{2}\right)={\tan }^2\theta$

$\Delta$ $ADE$ is isosceles
$\Delta$ $ABC$ is isosceles, $B=C$
$\tan \theta =\frac{AP}{DP}=\frac{\left(\frac{a}{2}\right)\tan B}{\left(\frac{a}{6}\right)}=3\tan B=3\tan C$
$\tan A=\tan ({180}^o-2B)=-\tan 2B=\frac{2\tan B}{{\tan }^{2}B-1}$
$=\frac{6\tan \theta }{{\tan }^2\theta -9}$
Since it is an isosceles triangle, $\cot \left(\frac{A}{2}\right)=\frac{AP}{BP}=\tan B=\frac{1}{3}\tan \theta \Rightarrow 9{\cot }^2\left(\frac{A}{2}\right)={\tan }^2\theta$