Geometry Problem

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

Click the figure below to see the complete problem 688.

## Wednesday, November 16, 2011

### Problem 688: Triangle, Angles, 10, 20, 30, 40, 60 Degrees, Measure, Mind Map, Polya

Labels:
20,
30 degrees,
60 degrees,
angle,
George Polya,
measurement,
mind map,
triangle

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http://img403.imageshack.us/img403/6706/sfvdf.png

ReplyDeleteExtend AB from B till F so that < EFD = 30°. Since < CEF = 50°, then < CDF = 50° and DF ⊥ BC, whereat < BDF = 30°, and finally from the cyclic quadrilateral CDEF we have x + 30° = 50° and x = 20°.

Extend AB till F so that AF=AC. This way DCFE is an isosceles trapezoid(CD=CF=EF and <DCF=<CFE=80). <AED =80=<EBD+<BDE. So <BDE=20.

ReplyDeleteDid you both used AD=AE which isn't given or how can someone conclude EF=CD?

ReplyDeleteReflection in the line BC. Let D→D'.

ReplyDelete∵ ∠DBC = ∠D'BC = 60°

∴ E, B, D' are collinear.

∵ ∠BDC = ∠BD'C = 80°

∴ ∠D'CA = 80°, ∠D'CB = 40°

Join DD'. Then BD = BD',

∴ ∠BDD' = 30°, ∠D'DC = 50°

∵ ∠D'DC = ∠D'EC = 50°

∴ C, D, E, D' are concyclic.

∴ ∠D'DE = ∠D'CE = 50°

∴ x = ∠BDE = ∠D'DE − ∠BDD' = 20°

Reflection in the line CE. Let D→D'.

ReplyDeleteThen ΔDCD' is equilateral.

∵ ∠DBC = ∠DD'C = 60°

∴ B, D, C, D' are concyclic.

∵ ∠EBD = ∠DCD' = 60°

∴ E, B, D' are collinear.

In ΔACD', ∠AD'C = 100°

∴ ∠ED'D = 40°

∵ ED = ED'

∴ ∠EDD' = 40°

∵ ∠BDD' = ∠BDD' = 20°

∴ x = ∠EBD = 20°

Find F on AB extended such that BC bisects < ACF. Then Tr. EFC is congruent with Tr. DFC and so EC = EF from which we can prove that Tr.s AFD and ACE are congruent. Hence AE = AD and so x = 20

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Or simply EF = FC = DC and CDEF is concyclic so x+30 = 10+40 hence x= 20

ReplyDeleteProblem 688

ReplyDeleteDraw the equilateral triangles DEK, then <EBD=<EKD=60.So EBKD is cyclic.Then <DEK=<DBK=60=<DBC.Therefore B,K,C are collinear. Is <EKD=60=2.30=2.<ECD,

Then the point K is circumcenter of triangle DEC .Then x=<EDB=<EKB=10+10=20.

MANOLOUDIS APOSTOLIS FROM GREECE

Here's my solution

ReplyDeletehttps://www.youtube.com/watch?v=6DCTArdbSpE

You draw a parallel to AC thro' B and then u draw a circle (D,DB) to intersect the parallel at say P.

DeleteYour proof then assumes that D,E,P are collinear -this is by no means evident or axiomatic. It must be PROVED.

Antonio -I would like u to pls comment too

Rgds

Sumith Peiris