In an equilateral triangle ABC, AD is perpendicular to BC. Find AD2 / CD2
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firstly, we know
AB=BC=CA
Take any∆
Lets go with ∆ ABD
in ABD
since AD is the perpendicular to BC it divides BC into two equal halves
BD=DC= 1/2 BC
AB²= AD²+BD²
AD²= AB²-BD²
= AB²-(1/2AB)²
= (AB+1/2AB)(AB-1/2AB)
= (3AB/2)(AB/2)
= 3AB²/4
in another ∆
CD²= AC²-AD²
therefore, AD²/CD²= 3AB²/4÷(AB²-3AB²/4)
= 3AB²/4 ÷( AB²/4)
= 3AB²/AB²= 3
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