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