Let $ABCD$ be a convex quadrilateral with $\angle BAC = \angle CAD$, $\angle ABC =\angle ACD$, $(AD \cap (BC =\{E\}$, $(AB \cap (DC = \{F\}$. Prove that: a) $AB\cdot DE = BC \cdot CE$ b) $AC^2 < \frac12 (AD \cdot AF + AB \cdot AE).$
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Tags: geometry, geometric inequality
Let $ABCD$ be a convex quadrilateral with $\angle BAC = \angle CAD$, $\angle ABC =\angle ACD$, $(AD \cap (BC =\{E\}$, $(AB \cap (DC = \{F\}$. Prove that: a) $AB\cdot DE = BC \cdot CE$ b) $AC^2 < \frac12 (AD \cdot AF + AB \cdot AE).$