In an equilateral triangle ABC, whose length of each side is 3 cm, D is the point on BC such that BD = ½ CD. What is the length of AD?

Given : AB = AC = BC = 3  cm and BD = $$\frac{1}{2}$$ CD

AE is median.

To find : $$AD = ?$$

Solution : BD + CD = 3

=> $$BD + 2BD = 3BD = 3$$

=> $$BD = \frac{3}{3} = 1$$ cm

Also, since AE is media => $$BE = CE = \frac{3}{2}$$ cm

=> $$DE = BE - DE = \frac{3}{2} - 1 = \frac{1}{2}$$ cm

Also, AE = $$\frac{\sqrt{3}}{2} a = \frac{3 \sqrt{3}}{2}$$ cm

In $$\triangle$$ ADE

=> $$(AD)^2 = (AE)^2 + (DE)^2$$

=> $$(AD)^2 = (\frac{3 \sqrt{3}}{2})^2 + (\frac{1}{2})^2$$

=> $$(AD)^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4}$$

=> $$AD = \sqrt{7}$$ cm