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
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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
https://photos.app.goo.gl/gLnATy32SQpvVVPu8
ReplyDeleteDraw DK perpendicular to BC
Note that BFDK is a rectangle and F, E, K are colinnear ( Simpson line)
so G is the midpoint of BD so OG perpen to BD