Given a convex quadrilateral $ABCD$ with $\angle BCD=90^\circ$. Let $E$ be the midpoint of $AB$. Prove that $2EC \leqslant AD+BD$.
Problem
Source: III Caucasus Mathematical Olympiad
Tags: geometry, geometric inequality, inequalities
18.03.2018 00:39
15.04.2018 20:13
Another way : Extend $BC$ to $X$ such that $ BC= CX$ , since $ \angle BCD$ is right , $BXD$ is isosceles $\Longleftrightarrow BD=DX$ and $AX=2EC$ Using triangle inequality on triangle $AXD$ we get $AX=2EC \leq AD+DX =AD+BD $
16.04.2018 19:26
bigant146 wrote: Given a convex quadrilateral $ABCD$ with $\angle BCD=90^\circ$. Let $E$ be the midpoint of $AB$. Prove that $2EC \leqslant AD+BD$. An extension (published here too): https://www.cut-the-knot.org/m/Geometry/ConvexDinca.shtml
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05.06.2018 21:49
Let $B'$ be the reflection of $B$ in $C$ and $C'$ be the reflection of $C$ in $E$.Now,$AC'$=$BC$=$B'C$,$AC$ common side and $\sphericalangle ACB'$=$\sphericalangle CAC'$,so $\bigtriangleup ACC'$ and $\bigtriangleup ACB'$ are congruent.Hence,$CC'$=$AB'$ and the conclusion follows by using triangular inequality on $\bigtriangleup ADB'$ with equality iff $A$,$D$,$B'$ collinear.