ABC is a triangle with a right angle at A. M and N are points on BC such that AM is the altitude, and AN is the angle bisector of ∠BAC. If CN/NB=21, what is CM/MB?

Answer 1

Note: Segment lengths BA and AB assumed to be equivalent in the following proof.

BN/sin(BAN)=BA/sin(BNA) (sine rule)
=> BA/BN=sin(BNA)/sin(45)

AC/sin(CNA)=NC/sin(45) (sine rule)
=> AC/NC=sin(CNA)/sin(45)

Since sin(BNA)=sin(180-BNA)=sin(CNA)

BA/BN=AC/NC, or
CN/NB =AC/AB=tan(B)

By metric relations
AM^2=BM*MC

By similar triangles,
AM/AB=MC/AC
=>AM=AB*MC/AC

Substitute in above
AB²*MC²/AC² = BM*MC
=> BM/MC = (AB/AC)^2
CM/MB=(AC/AB)^2=tan²(B)