Geometry Problem

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

Click the figure below to see the complete problem 830.

## Wednesday, December 12, 2012

### Problem 830: Quadrilateral, Triangle, Angles, 30 degrees, Congruence

Labels:
30 degrees,
angle,
congruence,
quadrilateral,
triangle

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http://img577.imageshack.us/img577/4899/problem830.png

ReplyDeleteSee drawing per attached sketch

Extend BC and from D draw DF ⊥to BC

We have ∠BDF=60 and DF=1/2. BD

In isosceles triangle ACD draw altitude CE

We have DE=1/2.AD=1/2.BD=DF

CE and CF are tangent lines from point C to circle centered D with radius=DE=DF ( see Sketch)

∠EDF=60+34=94

∠FCD=90-1/2. 94=43

So in triangle BCD, external angle FCD=30+x => x=43-30=13

Nice proof, Peter!

ReplyDeleteGeometric Solution of the problem 830 by Michael Tsourakakis from Greece

ReplyDeleteSee the image:http://img820.imageshack.us/img820/2807/geogebra.png

solution

CM is perpendicular of AD. Because CA=CD , CM is mediator of the AD.

AB and MC meet of G. So triangle AGD is isosceles .So , angle GDA=73

E is point of the GM with AED is equilateral triangle.

Angle GDE= angle EAG =73-60=13, therefore , angle EDB=73-34-13=26

Additional, angle DAB=73 and angle BDA=34 ,so , angle ABD=73,therefore DA=DB ,so ,DB=DE and triangle BDE is isosceles.Therefore,2 angle DBE=1800-260=1540 so, angle DBE= angle DEB=77.But angle CBD=30, therefore, angle EBC=47

But, angle BEC=77- angle CED=77-30=47.Therefore , triangle BCE is isosceles and CB=CE .But DB=DE .So , DC is mediator of the BE ,therefore, DE is bisector of the angle BDE=26.So,x = angle CDB=13

Draw CF and DE perpendicular to AD and BCrespectively, Then easily FD equal to DE and CFD and CED Tr.s are congruent.Hence < CDE = x + 34. so 2x + 34=60. So x=13.

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Problem 830

ReplyDeleteForm the equilateral triangle ADE ( the point C is inside the triangle ADE).So EC is bisector

Perpendicular the AD and <AEC=<CED=30(AC=CD, AE=ED) is <BDE=<ADE-<ADB=60-34=26.

But AD=BD=AE=ED(<ABD=180-73-34=73) then the point D is circumcenter the triangle ABE.

So <AEB=(<ADB)/2=34/2=17 and <BAE=(<BDE)/2=26/2=13. Then <EBC=<EBA-<CBA=(180-17-13)-(73+30)=47=<BEC (=17+30=47) therefore BC=CE.But BD=DE so triangleBCD=triangleECD.

Therefore x=<BDC=(<BDE)/2=26/2=13.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE