Geometry Problem

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

Click the figure below to see the complete problem 857.

## Tuesday, February 12, 2013

### Problem 857: Triangle, Median, Midpoint, Angles, 30, 45 Degrees, Congruence, Auxiliary Lines

Labels:
30 degrees,
45 degrees,
angle,
auxiliary line,
congruence,
median,
triangle

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CA ,perpendicular to AB. Then, angle ACE=60 ,and angle ECB=45.

ReplyDeleteBecause, angle EBC=45 ,EC=EB=ED=DA. So , 2x=30 ,x=15 degrees

http://img571.imageshack.us/img571/3688/p857trianglemedianmidpo.gif

Reflect C along AB to a point E.

ReplyDeleteAngle(EBC)=2*Angle(ABC)=2*(45)=90

By reflection, Triangle(AEC) is equiliteral and

since AD=DC, Angle(EDC)=90.

So, E,D,B,C are concyclic.

x = Angle(ABC)-Angle(DBC)

= 45 - Angle(DEC)

= 45 - (1/2)*(Angle(AEC))

= 45-30

= 15.

http://img42.imageshack.us/img42/6031/problem857.png

ReplyDeleteDraw CE ⊥AB

We have ∠ (ACE)= 60 and ED is a median of right triangle AEC

So ED=DA=DC

We have ∠ (ECB)=105-60=45 => EC=EB=ED

∠ (AED)=30=2.x =>x=15

This kind of problems usually are solvable by placing the circumcenter.

ReplyDelete- Let O be the circumcenter of (ABC).

- AD=OD=CD, so tr DOB is isosceles.

- tr OBC is equilateral.

- ODCB is a deltoid, so BD is perp to OC and <OBD=30°.

- <OBA=<OAB=15°.

Finally <x=15°.

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=35316&p=162751

ReplyDelete