In $\Delta ABC, AB = AC = \sqrt{3}$ and $D$ is a point on $BC$ such that $AD = 1$ . Find $(BD \times DC)$ .
What I Tried: Here is a picture :-
I have no idea for this problem. From the picture if $D$ lies a bit near of $B$ then I just reflect it to get $E$ near $C$ , $\Delta ADE$ is an isosceles triangle, but there's no use to it, how would that help in the original problem?
In fact, that did not look like the case when I tried it in Geogebra. $D$ lies nearly in the middle of $BC$ and, also with a bit of approximation, I got $(BD\times DC)$ to be nearly equal to $2$ , but why?
Can anyone help me with a reason here? Thank You.
Best Answer
Say $BC = 2x$. Say altitude from $A$ to $BC$ is $h$ and meets $BC$ at point $M$.
Given isosceles triangle,
$BM = MC = x$.
$h^2 = (\sqrt3)^2 - x^2 = 3 - x^2$
$DM = \sqrt {1- h^2} \,$
WLOG, assuming $D$ is left of $M$
$BD = BM - DM = x - \sqrt {1 - h^2}$
$CD = CM + DM = x + \sqrt {1 - h^2}$
$BD.DC = x^2 - 1 + h^2 = 2$