21. In a right-angled triangle, the angle bisector of an acute angle divides the opposite side into segments of length 1 and 2 . What is the length of the bisector?
A $\sqrt{2}$
B $\sqrt{3}$
C $\sqrt{4}$
D $\sqrt{5}$
E $\sqrt{6}$
An alternative solution can be obtained using the Angle Bisector Theorem. This gives us $A B: A C=B M: C M=1: 2$. Suppose, then, that $A B=x$ and $A C=2 x$. By Pythagoras' Theorem applied to triangle $A B C,(2 x)^{2}=x^{2}+3^{2}$. Therefore $x^{2}=3$. Therefore, by Pythagoras' Theorem applied to triangle $M B A$, $A M^{2}=x^{2}+1^{2}=3+1=4$. Therefore $A M=\sqrt{4}=2$.
21. Cabel the vertices of the right-angled triangle as $A, B, C$, with angle $A B C=90^{\circ}$. Let $M$ be the point where the angle bisector of $C A B$ meets the side $B C$. Let $N$ be the point where the perpendicular from $M$ meets the side $A C$. Let $x$ be the length of the bisector $A M$.
Then triangles $A B M$ and $A N M$ are congruent (they have all three angles the same, and one corresponding side $A M$ in common). Thus $M N=1$.
If we reflect triangle $M N C$ in the line $N C$, then we have an equilateral triangle with all sides equal to length 2 .
Hence $\angle N C M=30^{\circ}$ (half of $60^{\circ}$ ). Thus $\angle C A B=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$, and $\alpha=30^{\circ}$. Then triangles $M N C$ and $M N A$ are congruent (all angles the same, and common length $M N$ ), so $A M=M C=2$.