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5/6 <= AD/AB <= \sqrt{73}/10 , right triangle, <CAD=<ABE

Source: 2006 All-Ukrainian Correspondence MO of magazine ''In the World of Mathematics'', grades 5-11 p7

April 28, 2021
geometrygeometric inequalityright triangleUkraine Correspondence

Problem Statement

Let DD and EE be the midpoints of the sides BCBC and ACAC of a right triangle ABCABC. Prove that if ∠CAD=∠ABE\angle CAD=\angle ABE, then 56≤ADAB≤7310.\frac{5}{6} \le \frac{AD}{AB}\le \frac{\sqrt{73}}{10}.
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