## Thursday, January 21, 2010

### Problem 421: Right Triangle, Cevian, Angles, Measurement

Proposed Problem
Click the figure below to see the complete problem 421 about Right Triangle, Cevian, Angles, Measurement.

See more:
Complete Problem 421
Level: High School, SAT Prep, College geometry

1. http://geometri-problemleri.blogspot.com/2010/01/problem-63-ve-cozumu.html

2. AB=BCtan40
BD=ABtan20
EB=BCtan10
tan(EDB)=tan10/(tan20.tan40)=tan30
x=70-30=40
.-.

3. angle B + angle A + angle C =180
90 + 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,
hence,
x = 60

4. To Anonymous:

5. Why AB=BCtan40. Anybody can help.

6. Set A be Centre,
CED=30/2=15
CED+x=60
x=45

7. To Anonymous (March 19, 2010 10:55 PM):

8. Let circumcircle of EDC meet DA at F.
Triangle 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.

9. Pure Geometry proof

AD, 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

10. Problem 421
The 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