Geometry Problem

Click the figure below to see the complete problem 493 about Right Triangle, Circumcircle, Perpendicular, Chord.

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Complete Problem 493

Level: High School, SAT Prep, College geometry

## Friday, August 6, 2010

### Problem 493: Right Triangle, Circumcircle, Perpendicular, Chord

Labels:
chord,
circle,
circumcircle,
perpendicular,
right triangle

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Denote (XYZ) =angle XYZ

ReplyDeleteConnect AD and note that quadrilateral AFED is cyclic since (AFD)=(AED)=90

We have (EFD) = (EAD) ( angles face the same arc ED)

(EAD)=(CBD) (angles face the same arc CD)

(CBD)=(BDF)…………(corresponding angles)

So (EFD)=(BDF)

And ( DBF)=(BFG) (both angles complement to (EFD)=(BDF)

Triangles FGD and BGF are isosceles and GD=GF=GB

In isosceles triangle BOD , median OG will perpendicular to the base BD

Peter Tran

vstran@yahoo.com

We can use the simpson line !

ReplyDeleteAdil

ReplyDeleteYes, Simpson line is the better solution !

Thanks Adil.

< EFD = < EAD = < DBC = < BDF

ReplyDeleteHence in right triangle BFD, G is the centre and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Join AD and let m(DAC)=X

ReplyDelete=> m(ABD) = 90-X -----(1)

Since BFD is right triangle, m(BDF)=X ------(2)

Observe that AFED are concyclic => m(EFD) = m(EAD) = X ----------(3)

From (2) and (3) we can conculde that the triangle GFE is isosceles and GF=GD ---------(4)

Since m(BFD)=90 and from (4), it can be said that G is the circum-center of the right triangle BFD and hence BG=GD

Hence OG is perpendicular to the chord BD at its center G