Proposed Problem
Click the figure below to see the complete problem 421 about Right Triangle, Cevian, Angles, Measurement.
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Complete Problem 421
Level: High School, SAT Prep, College geometry
Thursday, January 21, 2010
Problem 421: Right Triangle, Cevian, Angles, Measurement
Labels:
angle,
cevian,
measurement,
right triangle
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http://geometri-problemleri.blogspot.com/2010/01/problem-63-ve-cozumu.html
ReplyDeleteAB=BCtan40
ReplyDeleteBD=ABtan20
EB=BCtan10
tan(EDB)=tan10/(tan20.tan40)=tan30
x=70-30=40
.-.
angle B + angle A + angle C =180
ReplyDelete90 + 50 + angle ECA + 10 = 180
hence, angle ECA = 30
now,
angle A + angle ECA + angle AEC = 180
50 + 30 + angle AEC = 180
angle AEC = 100
now,
angle EAD + angle AED + angle ADE = 180
20+100+angle ADE(or x)= 180
hence,
x = 60
To Anonymous:
ReplyDeleteThe answer is not correct.
Why AB=BCtan40. Anybody can help.
ReplyDeleteSet A be Centre,
ReplyDeleteCED=30/2=15
CED+x=60
x=45
To Anonymous (March 19, 2010 10:55 PM):
ReplyDeleteThe answer is not correct.
Let circumcircle of EDC meet DA at F.
ReplyDeleteTriangle FEC forms a trigonometric cevian model.
From angle C in clockwise, we get 60-x,10,10,100,30,x-30.
This is the classical 10-10-10-20-100-30 problem which makes x-30=10 => x=40.
Pure Geometry proof
ReplyDeleteAD, CE meet at F. Since AF = CF let MF the perpendicular bisector of AC meet BC at G and AB extended at H
Now BC is the perpendicular bisector of EH and since it therefore bisects < ECH and DF bisects < CFH (= 120) D is the incentre of Tr. FHC
Hence <DHG = 20, < BDH = BDE = 30 and so x must be < 40 since < ADB = 70
Sumith Peiris
Moratuwa
Sri Lanka
Problem 421
ReplyDeleteThe circle (D, DC) intersects the AC, CE in K, L respectively then <LDK=2<LCK=2.30=60.So
Triangle KDL is equilateral .Is <DLC=<DCL then <ELK=<EAK=50 so AKEL is cyclic. But <KLD=60=
2.30=2<KAD therefore the K is circumcenter the triangle AKD so LA=LK=LD=KD=DC and
<LAK=<LKA=50+30=80 or <LKE=<LAE=80-50=30.Then triangle LKE=triangle DKE so LE=ED and
<EDL=10,so <EDA=<BDA-<BDE=70-(20+10)=40.
Αnother solution the circle (E,EA) similar.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
It is special case of more general problem: (30-2t, 120-3t, 60-2t, 30) where t = 10.
ReplyDeleteThe problem is related with following cevian problem.
It can be generated by taking P on EC where AEDP concyclic.
https://output.jsbin.com/fofecum#100,10,10,30