Question

If triangle $ABC$ is an equilateral triangle such that $AD$ perpendicular $BC$ then $A{D}^2=k(DC{)}^2$then find the value of $k$

Answer 1

Finding the value of $k$

Given, $A{D}^2=k(DC{)}^2$

As the triangle is equilateral,
$$\begin{gathered} \Rightarrow AB=BC=AC \\ AD\perp BC \end{gathered}$$

According to the RHS congruency,

$$\begin{gathered} \Rightarrow \Delta ABD\cong \Delta ACD \\ \mathrm{as}AB=AC \end{gathered}$$

$AD=AD(\mathrm{common})$

$\angle ADB=\angle ADC=90°$

$$\begin{gathered} BD=CD=\frac{BC}{2} \\ BC=2CD \\ AC=2CD \end{gathered}$$

Applying the Pythagoras theorem.

$\begin{array}{rcl}AD²& =& AC²-CD²\\ AD²& =& \left(2CD\right)²-CD²\\ AD²& =& 3CD²\end{array}$

$k=3$

Hence, the value of $k$ is $3$ .