Inscribed angles PCD and PBA are equal, and vertical angles at P are equal making ∆ PCD ~ ∆ PBA. Then (PD/PA) = (DC/AB) = (2*EF/AB) = (EF/(1/2)AB) = (EF/AO).
Also, since Angle ADB is a right angle, ∆ ADP ~ ∆ AFE. Then (PD/PA) = (EF/AE).
Combining the two equations gives (EF/AO) = (EF/AE) or AO = AE.
m(DOC)=2a, then m(FAE)=a, m(FEA)=90-a
ReplyDeleteM is midpoint of DC, OM perpendicular DC. m(ODM)=90-a, m(DOM)=a
then
AFE=OMD
AE=AO
[URL=http://imageshack.us/photo/my-images/535/foxgeometry.png/][IMG]http://img535.imageshack.us/img535/5163/foxgeometry.png[/IMG][/URL]
ReplyDeleteThe two shaded right-angled triangles are congruent.
.: AE = OD = AO (radius of semicircle)
jyu
http://imageshack.us/photo/my-images/535/foxgeometry.png/
ReplyDeleteThe two shaded right-angled triangles are congruent.
.: AE = OD = AO (radius of semicircle)
jyu
Draw DB intersecting AC at P.
ReplyDeleteInscribed angles PCD and PBA are equal, and vertical angles at P are equal making ∆ PCD ~ ∆ PBA.
Then (PD/PA) = (DC/AB) = (2*EF/AB) = (EF/(1/2)AB) = (EF/AO).
Also, since Angle ADB is a right angle, ∆ ADP ~ ∆ AFE.
Then (PD/PA) = (EF/AE).
Combining the two equations gives (EF/AO) = (EF/AE) or AO = AE.