Geometry Problem

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

Click the figure below to see the complete problem 820.

## Monday, November 5, 2012

### Problem 820: Triangle, Cevian, Angles, 30 Degrees

Labels:
30 degrees,
angle,
cevian,
triangle

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analytical geometry and trigonometry (sorry)

ReplyDeletelet be A(0,0) C(1,0)

vectAD (ADcos86;ADsin86)

vectCE (-AEcos86-1;AEsin86)

the dot product of this two vectors is AD.CEcosx

x=66

I have a suggestion to make. Draw a line AG at 30 deg. to AC and let ED extended meet this new line in G. Now if anyone can prove that CAEG are concyclic then angle GEC is 30 deg which makes ang. AEC = 58 deg. and thus x = 58 + 8 = 66 deg.

ReplyDeleteFrom angle chasing, we have ∠ADC = 64, ∠EDA = 84.

ReplyDeleteReflect D along AC and becomes D’.

Note that E-A-D are collinear as ∠EAD + ∠DAC + ∠D’AC = 8 + 86 + 86 = 180.

Under reflection, DC = D’C and ∠DCA + ∠D’CA =2∠DCA = 60. So ΔDCD’ is equilateral.

Now connect DD” and observe ΔD’DE.

∠DED’= 180 – 92 = 88

∠EDD’= 84 + ∠ADD’= 84 + 64 - ∠D’DC = 84 + 64 – 60 = 88

So ΔD’DE is isosceles.

Combing the result, D’C = D’D = D’E. So D’ would be the circumcenter of ΔCDE.

∴∠DCF = ∠DCE = (1/2) ∠DD’E = (1/2)( ∠ADC - ∠D’DC) = (1/2)(64-60) = 2.

∴ x = ∠FDC + ∠DCF = 64 + 2 = 66.

Excelente. Muy creativa la solucion

ReplyDeleteExcelente. Muy creativa la solucion de W Fung. Gracias

ReplyDelete