Problem 2

In the triangle

ABC

, prove that there is a point

D

on side

AB

such that

CD

is the geometric mean of

AD

and

DB

if and only if

\sin A \sin B \le \sin^2 \frac{C}{2}

Step 1 of 5: Law of sines in the two sub-triangles

Previous step Next step

\frac{CD}{\sin A}=\frac{AD}{\sin C_1},\quad \frac{CD}{\sin B}=\frac{BD}{\sin C_2}

Detailed analysis

Let $D$ be a point on $AB$ , and set $C_1=\angle ACD$ , $C_2=\angle DCB$ , so $C=C_1+C_2$ . In $\triangle CAD$ , the angle at $A$ is $A$ itself (since $D\in AB$ ) and the angle at $C$ is $C_1$ , so the law of sines gives $\dfrac{CD}{\sin A}=\dfrac{AD}{\sin C_1}$ . Likewise, in $\triangle CDB$ the angle at $B$ is $B$ and the angle at $C$ is $C_2$ , giving $\dfrac{CD}{\sin B}=\dfrac{BD}{\sin C_2}$ .