See complete Problem 156

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

Post your solutions or ideas in the comments.

## Tuesday, August 5, 2008

### Elearn Geometry Problem 156

Labels:
chord,
circumcircle,
circumradius,
excircle,
exradius,
secant,
similarity,
triangle

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I have first half of solution, but on the second part

ReplyDeleteusing 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 )

Please verify

To c.t.e.o:

ReplyDeleteYour conclusion 2R.r1 = AD.PE is OK.

Other conclusions:

AD.PE = AE.DE = 2R.r1

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

ReplyDeleteDEG = 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

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

CD/AD = PD/CD

DE/AD = PD/DE ( CD=DE)

(AE - AD)/AD = (PE - DE)/DE

AE/AD - 1 = PE/DE - 1

AE/AD = PE/DE

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

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

ReplyDeleteSolution to problem 156.

ReplyDelete1) 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.