Probability that a side length is greater than 2 In triangle $ABC$, we have $\angle B=60^\circ$, $\angle C=90^\circ$, and $AB=2$.
Let $P$ be a point chosen uniformly at random inside $ABC$. Extend ray $BP$ to hit side $AC$ at $D$. What is the probability that $BD<\sqrt 2$?
I drew a picture for this but now I am stuck on how to continue.
 A: I think this problem is actually quite interesting and straightforward.
Clearly, triangle ABC is a 30-60-90 triangle, giving us leg lengths of $1$ and $\sqrt{3}$. Construct BD such that $BD=\sqrt{2}$. We have created a 45-45-90 triangle nested in the big 30-60-90 triangle.
Realize that if P is above the segment BD, BD will surely be greater than $\sqrt{2}$. If it's under segment BD, then BD will be less than $\sqrt{2}$. We arrive at the conclusion that if P is in triangle BDC, BD is less than $\sqrt{2}$.
How do we find the probability? Use area. The probability that P is in triangle BDC is just the area of triangle BDC divided by the area of triangle ABC. The area of triangle ABC is $\frac{\sqrt{3}}{2}$ and the area of triangle BDC is $\frac{1}{2}$. Then we divide: $$\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\boxed{\frac{\sqrt{3}}{3}}$$
We can also just use the ratio of CD to DA here because the heights of the two triangles ABC and BDC are equal. $\frac{CD}{DA}=\frac{\sqrt{3}}{3}$.
Hope this helped.
