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 1028.

## Saturday, July 12, 2014

### Geometry Problem 1028: Right Triangle, Double Angle, Midpoint, Congruence

Labels:
congruence,
double angle,
Geometry,
mathematics,
midpoint,
Problem,
right triangle

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Solution:

ReplyDelete1)Connect B and E Notice that Triangle ABE is isosceles.

2)Let AE = AB = a, let DE = BE = EC = b(right triangle).

3)AEB ~ DBE (They both isosceles and have the same angles). From here we get that AE/BE = BE/DB =

a/b = b/DB --> DB = b^2/a.

4)Draw EN_|_BC and EM _|_AC (notice that EN = DB/2 = b^2/2a).

5)AEM ~ CEN (Same angles) --> AE/EM = EC/ EN = a / EM = b/(b^2/2a).

6)From the proportion in step 5 we get that EM = b/2, in triangle ECM angle ECM = 30 = 90 - 4x.

Solution : The measure of x is 15.

Trigonometry solution:

ReplyDeleteConnect BE and let F is the midpoint of BE

BDE and BAE are isosceles and similar

Let BC= 1we have EC=EB= 1/(2.cos(x)) and BH= 1/(4.cos(x))

And AB= BF/sin(x)= 1/(2.sin(2x))

AC= AB/cos(3x)=BC/sin(3x) => tan(3x)=2. sin(2x)

This equation has 2 solutions: x= 0 and x=15

Prob 1027

ReplyDeleteWe 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 New Jersey

construim AN_|_BE si EM _|_AC, N pe BE,M pe BC,avemAEB ~ DEB(EM=NE=EB=BE/2=CE/2=>ECA=30, 90-4X =30=> X =15

ReplyDeleteWonderful solution, Prof. Radu. I wish your solutions were in English, though, for Google translation is never adequate or accurate.

DeleteTrigonometric Solution 999 and 1028 ( Because they have the same picture and same solution

ReplyDelete1. EF perpendicular AB ---> BC=2EF

2. Triangle AEF EF= ABsin2X

3 Triangle ABC BC= ABtg3X

Tg3X=2sin2X From geometric conditions 0 <3X < 90 ---> 0 <X<30

Equation has this solution X=15

Erina New Jersey

Hi Antonio

ReplyDeleteWe have 2 other geometric solutions for problems 1028 .

If you want them I can send to you.

Have a great night .

Erina

Hi Erina,

DeleteGreat. Please send your geometric solutions. Thanks

Let AE = AB = q, BD = r and BE = DE = CE = p

ReplyDeleteBE is tangential to ADE so p^2 = rq ....(1)

From similar Tr.s EF altitude of Tr. AEF = qr/2p = p/2 from (1)

So < ECA = 30 and hence x = 15

Sumith Peiris

Moratuwa

Sri Lanka

E is circumcenter of DBC => m(ABE)=90-X => ABE is isosceles triangle.

ReplyDeleteLet AP be the altitude of ABE , EQ be perpendicular from E onto AC.

APE and AQE are congruent

=> EQ=EP=BE/2=EC/2

=> EQC is 30-60-90 right triangle

=> m(QCE)=90-4X=30

=> x=15