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[JBMO 2015 No 3] Banyak tegak lurusnya

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$ABC$ segitiga lancip. Garis $l_1$ dan $l_2$ tegak lurus terhadap $AB$ masing-masing di titik $A$ dan titik $B$. Garis dari $M$, titik tengah $AB$, yang tegak lurus dengan $AC$ dan $BC$ masing-masing bertemu $l_1$ dan $l_2$ di $E$ dan $F$. Jika $EF$ dan $CM$ bertemu di $D$, buktikan $$\angle{ADB}=\angle{EMF}$$


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Misal $ME$ memotong $AC$ di $P$ dan $MF$ memotong $BC$ di $Q$.



Karena $MQ . MF = MB^2 = MA^2 = MP . ME$ Maka $EPQF$ segiempat talibusur. Dan karena $CPMQ$ juga segiempat talibusur akibatnya $\angle MCP = \angle PQM = \angle FEM$ sehingga $EPDC$ ialah segi empat talibusur.

Lebih lanjut diperoleh $\angle EDM = 90$. akibatnya $AMDE$ juga segiempat talibusur. Sehingga $\angle DAM = \angle DEM$. analog $\angle DBM = \angle DFM$. Jadi terbukti $\angle ADB = \angle EMF$

Edited by Louiscahyadi
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1. itu $FAM$ atau $FEM$? ._>


2. spoiler anda =.=


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