Geometry 1

In $\triangle ABC$ , let $D\in CA$ and $E\in AB$ such that $\angle ABD=\angle CBD$ and $\angle ACE=\angle BCE$ . A rhombus is inscribed into the quadrilateral $BCDE$ (that is, the vertices of the rhombus lie on distinct sides of $BCDE$ ). Let $\phi$ be the non obtuse angle of the rhombus. Prove that, $\phi\le \max(\angle ABC,\angle ACB)$ .

This problem belongs to the set TST Bangladesh 2014 Problems.

#Geometry #Inequalities #Angles #Quadrilaterals #Rhombus

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$