Question

In $△ABC$, where $A,B,C$ are acute, the distance of the orthocenter from the sides are in the proportion

• A. $\cos A:\cos B:\cos C$
• B. $\sin A:\sin B:\sin C$
• C. $\sec A:\sec B:\sec C$
• D. $\tan A:\tan B:\tan C$

Answer 1

The correct option is C $\sec A:\sec B:\sec C$

Here, $AD,BE,CF$ are the altitudes

Also, $H$ is the orthocenter

$\angle HCD=90-B$ [$\because \angle HDC={90}^o$]

$\therefore \angle DHC=\angle B$

Now, in $△CDH$

$\sec B=\frac{HC}{HD}$.......(1)

similarly,

in $△CHE$

$\angle HCE=90-A$

$\angle CHE=A$

$\sec A=\frac{HC}{HE}$.......(2)

from (1) and (2)

$\frac{HD}{HE}=\frac{\sec A}{\sec B}$ .....(3)

Similarly,

$\frac{HE}{HF}=\frac{\sec B}{\sec C}$.......(4)

from (3) and (4)

$\Rightarrow HD:HE:HF=\sec A:\sec B:\sec C$