Proposed Problem

Click the figure below to see the complete problem 337 about Isosceles Trapezoid inscribed in a circle, Angle bisector, Parallel, Concyclic points.

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Complete Problem 337

Level: High School, SAT Prep, College geometry

## Saturday, August 15, 2009

### Problem 337. Isosceles Trapezoid, Angle bisector, Parallel, Concyclic points

Labels:
angle bisector,
concyclic,
cyclic quadrilateral,
isosceles,
parallel,
trapezoid

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ang ADC + ang BCD = 180 (from ang of one side in trap)

ReplyDeleteso need to prove ang EDH(C) + ang EGH = 180

ang EDC = ang ADC + ADE (or ECD), (CE bisector)

so for ang EDC have to add ECB ( BCD - ECD )

but ECB = EBC ( equal bisector for equal ang )

so ang EGH = ang EBC (are at the same side of L1 //BC)

http://s17.postimg.org/p2m3e2le7/pro_337.png

ReplyDeleteConnect DE , EG and EB

Since E is the midpoint of arc AD and trapezoid ABCD is isosceles

So EBC and EGF are isosceles triangles.

And ∠ (GFE)= ∠(FGE)= ∠(BCE)= ∠(EBC)

So B, G, E are collinear

BCDE is cyclic => ∠(EDC) supplement to ∠(EBC)=> ∠(EDC) supplement to ∠(FGE)

So EGHD is cyclic