Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 759 details.

## Friday, June 1, 2012

### Problem 759: Equilateral Triangle, Transversal, Trisection of sides, Congruence, Angle

Labels:
angle,
congruence,
equilateral,
transversal,
trisection

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By Menelaus theorem on triangle ABC, (BD/DC)*(CF/FA)*(AE/EB)=1.

ReplyDeleteHence CF/FA = 4/1. Therefore FA:AC=1:3.

Since ABC is an equilateral triangle, FA = (1/3)AC = (1/3)AB = AE

So triangle AEF is an isosceles triangle. Angle EAF = 120, hence x = 30.

please disregard my previous solution. new Solution below include minor typo corrections.

ReplyDeleteApply Menelaus’s theorem to triangle ABC with secant FED

DB/DC x FC/FA x EA/EB=1 => ½ x FC/FA x ½= 1

So FA/FC= ¼ => CA/CF= ¾

From A draw AH //FD ( H on CB)

We have CA/CF=CH/CD= ¾ => CH=3/4.CD=3/4 .2/3 .CB= ½. CB

AH will be an angle bisector and altitude of triangle ABC, so x= angle(HAC)= 30

Draw EG ∥ DC (G on AC)

ReplyDelete∆s GDC and AEG are equilateral and

BEGD is a parallelogram

1/2 = EG/DC = FG/FC

G is the midpoint of FC

FG = GC = EB = DG

∆FDC is rt angled at D

x = 30°

AE = BD = a/3 where AB = BC = AC = a

ReplyDeleteSo BE = CD = 2a/3

Apply Menelau's Theorem to ∆ABC

and transversal DEF

(BE/EA).(AF/FC).(CD/DB) = 1 numerically

Easy to see BE/EA = 2 = CD/DB

Follows AF/FC = 1/4, AF = a/3 = AE

So x = ∠AFE = ∠AEF = (1/2)∠EAC = 30°

Let G be the midpoint of [BE].(Draw a picture yourself)

ReplyDeleteFrom BG/BA=BD/BC =1/3 follow that triangles BDG and BAC are similar; moreover DG/CA=1/3.

Now, in triangle BDE, median DG=BE/2- half of side of base, or that suffice that angle BDE is right. In fact, because triangle BDG is equilateral, follow that triangle EBD (in that order of vertices) is of type 30-60-90.

The rest is simple.

Visit also http://ogeometrie-cip.blogspot.com/

Let BD = AE = a . Draw DG // AB intersecting AC at G. now DC = BE = 2a. In tr.ABC, by BPT, GC = 2a and AG = a.

ReplyDeletenow tr. DGC is equilateral. so DG = 2a . In tr. FGD and tr.FAE are similar. so FA = a. Now FA = AE = a. So ∠FEA = ∠AFE = x. ∠ FEA = ∠ GDF = x. In tr. DFG ∠DFG + ∠ GDF = ∠DGC. 2x = 60° or x = 30°

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27308&p=133642

ReplyDeleteEasily BED is a 30-60-90 Tr. Having BE = 2.BD and the included angle 60. So < BED = 30 and hence x = 30.

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka