2 parallelograms, equal segments, double segments, equal angles wanted

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2 parallelograms, equal segments, double segments, equal angles wanted

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Source: Euler Olympiad 2018 P8 [junior]

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#1 h Sep 9, 2018, 7:04 PM • 1 Y

Y by Adventure10

Vertex $F$ of a parallelogram $ACEF$ lies on the side $BC$ of a parallelogram $ABCD$ . It is known that $AC = AD$ and $AE = 2CD$ . Prove that $\angle CDE = \angle BEF.$

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#2 h Sep 11, 2018, 11:29 AM • 2 Y

Y by Maths_Guy, Adventure10

Let $\angle ADC = \alpha$ and $O=AE\cap CF$ . Also let $\angle ADE = \delta$ and $\angle FEB = \theta$ .

We have $ADCO$ is a isosceles trapezium. As $ADCB$ is parallelogram, $AB = AO$ . We have from law of sines in $\triangle AEB$ , $\frac{1}{2}=\frac{AB}{AE}=\frac{\sin (3\alpha + \theta - 180^{\circ})}{\sin (180^{\circ}-\alpha - \theta)}= -\frac{sin (3\alpha + \theta)}{\sin (\alpha + \theta)}.$ Suppose $\gamma = \alpha + \theta$ . Thus $\frac{\sin (2\alpha + \gamma)}{\sin \gamma} = -\frac{1}{2}$ . Thus $\cot \gamma = \frac{-1-\cos 2\alpha}{\sin 2\alpha}.$
Notice that $\delta = \alpha + \angle CDE$ , thus we need to show that $\gamma = \delta$ . Now law of sines in $\triangle AED$ , $\frac{\sin (\alpha +\delta)}{\sin \delta} = \frac{AD}{AE}=\frac{AC}{2AO}=\frac{1}{2}\cdot \frac{\sin \alpha}{\sin 2\alpha}.$ Thus we have $\cot \delta = \frac{\frac{1}{2}\cdot \frac{\sin \alpha}{\sin 2\alpha}-\cos \alpha}{\sin \alpha}=\frac{1}{2\sin 2\alpha} - \cot \alpha.$
Thus we need to show $-1-2\cos 2\alpha =1 - 2\sin 2\alpha \cot \alpha = 1 - 4\cos ^2 \alpha$ or $2 = 4\cos ^2 \alpha - 2\cos 2\alpha = 4\cos^2 \alpha - 2(2\cos ^2\alpha -1) = 2$ which is true.

Hence Proved

This post has been edited 1 time. Last edited by TheDarkPrince, Sep 11, 2018, 11:29 AM

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#3 h Sep 11, 2018, 1:15 PM • 3 Y

Y by nguyendangkhoa17112003, Adventure10, Mango247

Let $M=AE\cap BC$ , i.e. $AM=ME=CD=AB$ , thus $\triangle ABM\sim\triangle CAB\implies\ AB^2=BM\cdot BC \ (\ *\ )$ , i.e. $ME^2=BM\cdot MD\ (\ 1\ )$ , $AMCD$ is an isosceles trapezoid, wherefrom $\angle MDC=\angle MAC=\angle MEF\ (\ 2 \ )$ , $\angle AMB=\angle ADC=\angle ACD=\angle AMD\ (\ 3\ )$ . From $(1)$ and $(3)$ we get $\triangle BME\sim\triangle EMD$ and subsequently $\angle BEM=\angle EDM$ ; with $(2)$ we are done.

Best regards,
sunken rock

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#4 h Sep 12, 2018, 8:16 PM • 2 Y

Y by Adventure10, Mango247

$\begin{gathered} {\text{CN// = AB}}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \left\{ \begin{gathered} {\rm B}{\rm N}// = {\text{EF}} \hfill \\ {\rm A}{\rm B}{\rm N}{\text{D}}\mathop {}\limits_{}^{} {\text{ }}{\text{,}}\mathop {}\limits_{}^{} {\text{AEND}}\mathop {}\limits_{}^{} {\text{isosceles}}\mathop {}\limits_{}^{} {\text{ trap}}{\text{.}} \hfill \\ \end{gathered} \right.\mathop {}\nolimits_{}^{} \Rightarrow \hfill \\ {\rm A},{\rm B},{\rm E},{\rm N},{\text{D}}\mathop {}\limits_{}^{} {\text{are}}\mathop {}\limits_{}^{} {\text{concyclyc}}\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} {\text{F}}\widehat {\text{E}}{\text{B = }}\sphericalangle {\text{E}}\widehat {\text{B}}{\text{N = }}\sphericalangle {\text{E}}\widehat {\text{D}}{\text{N}} \hfill \\ \end{gathered}$

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