In triangle ABC, AD is drawn perpendicular from A on BC. If AD^2 = BD.CD................Then angle BAC is

In triangle ABC, AD is drawn perpendicular from A on BC. If AD^2 = BD.CD................Then angle BAC is?????

let angle BAD = x , DAC = y => x + y = angle BAC
tan x = BD/AD => AD = BD/tanx
tan y = CD/AD => AD = CD/tan x
AD^2 = (BD * CD)/(tan x * tan y)
AD^2 = BD * CD => tan x * tan y = 1
tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)]
tan(x + y) = infinity
BAC = x + y = pi/2 or 90 degrees

...a right angle.

More explanation:
AD^2 = BD.CD can be interpreted as CD/AD = AD/BD meaning Triangle CAD is proportional (similar) to triangle DAB. This means angle CAD = angle ABD and angle DCA = angle DAB. In the triangle ABC the angle sum is then 2(CAD+DAB) therefore CAD+DAB = CAB = 90 degrees.