MathLabs

Problem 2

In the triangle ABCABC, prove that there is a point DD on side ABAB such that CDCD is the geometric mean of ADAD and DBDB if and only if sin⁡Asin⁡B≤sin⁡2C2\sin A \sin B \le \sin^2 \frac{C}{2}.
Step 4 of 6: Turning the height comparison into an area comparison
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CF≤MN ⟺ 12CF⋅AB≤12MN⋅AB ⟺ [ABC]≤[MAB]CF\le MN\ \Longleftrightarrow\ \tfrac12CF\cdot AB\le \tfrac12MN\cdot AB\ \Longleftrightarrow\ [ABC]\le[MAB]
Detailed analysis

Multiplying both sides of CF≤MNCF\le MN by 12AB\tfrac12AB turns the height comparison into a comparison of areas: since CFCF and MNMN are the heights from CC and MM onto the common base ABAB, this says [ABC]≤[MAB][ABC]\le[MAB].