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 2 of 5: Multiplying the two relations

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CD=AD\cdot\frac{\sin A}{\sin C_1},\quad CD=BD\cdot\frac{\sin B}{\sin C_2}\ \Rightarrow\ CD^2=AD\cdot BD\cdot\frac{\sin A\sin B}{\sin C_1\sin C_2}

Detailed analysis

Solving each relation for $CD$ and multiplying them gives $CD^2=AD\cdot BD\cdot\dfrac{\sin A\sin B}{\sin C_1\sin C_2}$ . Hence $CD^2=AD\cdot BD$ holds exactly when $\sin A\sin B=\sin C_1\sin C_2$ for the particular split $C=C_1+C_2$ determined by $D$ .