## Thursday, July 10, 2014

### Geometry Problem 1027: Triangle, Double, Quadruple, Angle, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1027.

1. Let b= AC=AD
We have DC=DE= 2. b. sin(2α )
∠(BED)=3α
Relation in triangle AED => ED/sin(2α )= AD/sin(3α) = 2.b.sin(2α )/ sin(2α )= 2.b
So sin(3α )= ½= sin(30) => α = 10
So x= 90- 4α = 50

2. There is a geometric solution for this problem ,right?

1. To Dima
Problem 1027
Yes, there is a geometric solution, Thanks.

3. We build
AH perpendicular BC
DF perpendicular AB

DF=DH = 1/2ED X =50 grade

Erina New Jersey

1. Hi Erina,
Please, send again your complete solution, there is some problem with the system.
Thanks. Antonio

4. Hi Antonio

Please publish the full solution

Prob 1027
We build AH perpendicular BC and DF perpendicular AB
CH=HD=1/2DC=1/2ED
DF=DH=1/2ED
α =10
X=90-4α
X=50

Prob 1028

BE=ED=EC EM=EN EN =1/2 EC
Triangle ECN < ECN =30 ECN = 90-4X -> X =15

Prob 1029
We note <BDF = 90
BD perpendicular AC

Erina Erina Jersey

5. We build

1. AH perpendicular BC --> CH =HD =1/2 DC = 1/2ED
2. DF perpendicular AB ----> DF=DH =1/2 ED
3. triangle DEF α =10
4. X=90-4α X = 50

Erina New Jersey