Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 802.
Monday, September 3, 2012
Problem 802: Triangle, Median, Perpendicular, Angle, Measurement
Labels:
angle,
measurement,
median,
perpendicular,
triangle
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Prolongar el segmento ED, desde D, 2 unidades hasta el punto que llamamos F, de modo que el segmento EF mide 4 y por tanto el triángulo rectángulo CEF es isósceles, y así el ángulo CFE=45º. Como los triángulos CDF y ADE son congruentes, por ser CD=AD, DF=DE y ángulos opuestos por el vértice en D, se tiene que ángulo AED =àngulo CFD, esto es x=45º.
ReplyDeleteMigue.
tan x
ReplyDelete= (AE sin x) / (AE cos x)
= (AD sin ADE) / (DE - AD cos ADE)
= (DC sin CDE) / (DE + DC cos CDE)
= (4) / (2 + 2)
= 1
x = 45 degrees
By Pyth. Thm and median, we have AD = DC = sqrt(20) = 2*sqrt(5).
ReplyDeleteSimply let angle EDC be Y.
Hence, using sine rule,
(2*sqrt(5))/(sinX) = 2/(sinEAD)
= 2/(sin(Y-X))
= 2/(sinYcosX - sinXcosY)
Note that sinY = 2/sqrt5 and cosY = 1/sqrt5
=> 2cosX - sinX = sinX
=> cosX = sinX
X = 45.
Extend the median to a point P such that AP is perpendicular to EC. Since AD = DC, triangle PAD is congruent to triangle ECD.
ReplyDeletetanX = PA/PE = 4/(2+2) = 1, X = 45
It should be AP perpendicular to BD, parallel to EC.
DeleteYes , thanks Jacob
Deletethanks for correction, jacob
DeleteWe draw AP⊥BD
ReplyDelete1.ΔAPD= ΔDEC =>ED=DP=2
AP=EC=4
2.Δ APE, AP=4=EP=>angle AED=45º
Another solution
We draw AF⊥CE=>AF‖DF
AD=DC=>EF=FC=4
AF=2ED=4
ΔAEF, angle AEF=45º=>angle AED=45º
ERINA.H NJ
Note that AD/AC=ED/EC=1/2
ReplyDeleteSo AE is the external angle bisector of angle DEC
and x=1/2.90=45
Peter Tran
Draw DF parallel to CE (F on AE)
ReplyDeleteSo FDE = DEC is a right angle
DF = half CE = 2 = DE
cot x = DE/DF = 1
x = 45 deg
http://www.alnasiry.com/upload//uploads/images/domain-d4dc4fd216.png
ReplyDeletedraw ED and MD=2 then AECM is parall.
AM=ME and M<EMA=90 then x=45
Let line perpendicular to AC that passes through A meet BD at P. PA/AD=4/2=PA/(2*sqrt(5)), so PA=4*sqrt(5) while AC=2*(2*sqrt(5))=PA, because sqrt(4^2+2^2)=2*sqrt(5). Then <ACP=45=<AEP.
ReplyDeleteExtend ED to F such that DF=2. Then AECF is a parellogram, EFC is an isoceles right Tr and so x=45
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
Problem 802
ReplyDeleteSuppose that M is the medium of EC, then EM=2=ED.So <EDM=45.But AD=DC then AE//DM .Therefore x=<AED=<EDM=45.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Problem 802 (sοlution 2)
ReplyDeleteLet AN//BD ,or AN perpendicular CE (N belongs to the extension of CE).Then NE=NC=4
And AN=2.ED=2.2=4=NE.So <NEA=45=<AED=x.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
CD=sqrt20=AD
ReplyDeletetan<EDC=2
<EDC=arctan(2)=63.43494882
<EDA=116.5650512
AE^2=AD^2+2^2-2(2)(AD)cos<EDA
AE^2=32
cosx=(AE^2+ED^2-AD^2)/2(AE)(ED)
x=45