Let $\triangle ABC$ be an isosceles.$(AB=AC)$.Let $D$ and $E$ be two points on the side $BC$ such that $D\in BE$,$E\in DC$ and $2\angle DAE = \angle BAC$.Prove that we can construct a triangle $XYZ$ such that $XY=BD$,$YZ=DE$ and $ZX=EC$.Find $\angle BAC + \angle YXZ$.
Problem
Source: Balkan MO 2014 G-3
Tags: geometry
mathmatecS
10.06.2015 18:06
$\angle BAD+\angle CAE=\angle DAE\cdot\cdot\cdot a$ So we can fold $\triangle ABD$ by $AD$, $\triangle AEC$ by $AE$ and we can name folded triangle $\triangle ADB'$, $\triangle AEC'$. By $a$, two triangle touches and B' and C' is a same point. Therefore $\triangle DEB'$ can be create. So $\angle BAC$ + $\angle YXZ$ = $\angle BAC$ + $\angle DB'E$ = $\angle BAC$ + $\angle DB'A$ + $\angle EB'A$ = $\angle BAC$ + $\angle ABC$ + $\angle ACB$ =180
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kimcj000310
10.06.2015 20:04
Really nice solution!!!
samithayohan
10.06.2015 21:23
My solutionis very similar to mathmatecS solution. There is nothing to do after taking the reflections from $AD$ and $AE$
mathmatecS
11.06.2015 11:59
But idea of reflecting is very important