Tuesday, August 5, 2008

Elearn Geometry Problem 156

See complete Problem 156
Triangle, Circumradius, Exradius, Chord, Secant line. Level: High School, SAT Prep, College geometry

1. I have first half of solution, but on the second part
using similarity of tr AOE, (OE perpendicular to AD)
and tr PEG, ( P AD meet BC, G tg point of E to BC )
I get 2R∙r = AD∙PE, ( ang EAO = DEG = 90 -x - C )

2. To c.t.e.o:
Other conclusions:

3. x = A/2, T tg point of E to AC

DEG = 90 - x - C ( first comment )
DEC = DEG + GEC
DEC = 90 - x - C + C/2 ( CE bisector of GCT )

DEC = 90 - x - C/2 (1)

ECT = ECG = 90 - C/2
DCE = GCE - GCD

DCE = 90 - C/2 - x (2)
( GCD = x ,have same arc DC as A/2)

from (1) & (2)
DEC = DCE
=>

DE = DC
------------------------------------------
▲ADC ~ ▲PCD ( A/2 = PCD = x , see above)

AE/AD - 1 = PE/DE - 1

AE∙DE = AD∙PE = 2R∙r (see first comment)
-----------------------------------------

4. http://ahmetelmas.wordpress.com/2010/05/15/geo-geo/

5. Solution to problem 156.
1) The bisector of ang(ABC) meets AE at point I, and the circumcircle at M. Arcs AM and MC are equal. As AE is bisector of ang(BAC) then arcs BD and CD are equal. BE is external bisector, so ang(EBM) = 90º.
We have ang(DBE) = 90º - ang(MBD) = 90º - arc(MD)/2.
Besides ang(DEB) = 90º - ang(BID) = 90º - (arc(AM) + arc(BD))/2 = 90º - (arc(MC) + arc(CD))/2 = 90º - arc(MD)/2 = = ang(DBE). Hence BDE é isosceles, with BD = DE.

2) The extension of BO meets the circumcircle at F, and ang(BDF) = 90º. Let T be the tangent point of the excircle with AC. We have ang(BFD) = ang(BAD) = ang(EAT). So triangles BDF and ETA are similar, with AE/FB = ET/BD. Therefore AE/(2R) = r1/DE and AE.DE = 2Rr1.