Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it

Example 5. As shown in Figure 5, given that $\mathrm{ABCD}$ is a cyclic quadrilateral, and diagonal $A C$ bisects $B D$ at $E$ , prove that $A B^{2}+$ $\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{DA}^{2}=2 \mathrm{AC}^{2}$ .

Solution

Prove $\because \mathrm{S}_{\triangle \triangle D C}=\frac{1}{2}$ .
$\mathrm{AC} \cdot \mathrm{DE}_{\sin \mathrm{\alpha}}, \mathrm{S}_{\Delta \Delta \mathrm{Bc}}=\frac{1}{2}$ .
$A C \cdot B E \sin \alpha, \therefore S_{\triangle A B C}=S_{\triangle A D C} \Longrightarrow A D \cdot C D=$ $\mathrm{AB} \cdot \mathrm{BC}$ .

Also, $\mathrm{AC}^{2}=A B^{2}+B C^{2}-2 \mathrm{AB} \cdot \mathrm{BC} \cos \angle \mathrm{ABC}=$ $\mathrm{AD}^{2}+\left(\mathrm{D}^{2}+2 \mathrm{AD} \cdot \mathrm{DCco} \angle \mathrm{ABC} \Longrightarrow 2 \mathrm{AC}^{2}=\mathrm{AB}^{2}\right.$ $+B C^{2}+C D^{2}+\mathrm{DA}^{2}$ , hence proved.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.