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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 691.
This is far easy one!AC=CB and AE perpendicular to DC, then ACED is a deltoid so <CDE=30.x=50-30=20 and we are done here (:
http://img191.imageshack.us/img191/4097/problem691.pngSince triangle ABC is isosceles so (BAC)=(BCA)=80And (DAE)=40 and (DCE)=30Triangle DAE congruent to Tri. CAE ( case SAS)So DE=EC and ( CDE)=(ECD)=30In triangle DAF, x=180-80-80=20Peter Tran
Sí pero yo creo que quisiste decir que AC = AD o no?
AC=AD, efectivamente. gracias @Markov
<ADC is an exterior angle, then tr ADC is isosceles, so we have AD = AC, cause this triangles CAE and DAE are congruents by SAS.Then ADF is 80 80 20 , therefore x=20 Greetings ^^
AE is easily seen to be the perpendicular bisector of DC hence Tr. DEC is isoceles and < CDE = 30 and so x= 20Can anyone show that Tr. BEF is isoceles? Sumith PeirisMoratuwaSri Lanka
Using symmetry, <CEF and <DEB are 60 each. Since <BDE is 100 and <CEF is 100 by difference (180- 60-20). Therefore, TR DEB is congruent to Tr ECF (ASA). Since corr sides of cong Tr are equal, EF = EB making TR BEF isoceles.
....and that BD = CF?
...also that AE is perpendicular to BF?