Geometry Problem

Click the figure below to see the complete problem 448 about Tangent Circles, Diameter, Perpendicular, Chord.

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Complete Problem 448

Level: High School, SAT Prep, College geometry

## Saturday, May 8, 2010

### Problem 448: Tangent Circles, Diameter, Perpendicular, Chord

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DCE = DAB = 1/2 DO'E = DOB ( DO'E, DOB isoc, ODB comm )

ReplyDeleteDEC = DCA ( have same arc DC )

=>

▲ADC ~ ▲DCE (1)

DC common for ADC, DCE (2)

(1) & (2)

▲ADC = ▲DCE

=>

AD = DC = DE = x (3)

▲DGE ,

GDE = 90 - ADG = 90 - B

GED = 90 - B (ADC isoceles => DG bisector )

=>

DGE isoceles =>

▲DGE ~ ▲ADC

=>

x/a = b/x

x² = ab

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Tr.s ADC and DCE are not necessarily congruent. This will be so only in the special case of each triangle being isoceles.

DeleteCould u elaborate?

consider common tangent, mBAD=mECD

ReplyDeletehence mADC=mBCE=mCDE=pi/4, mEGD=pi/4 as well

draw DH//AB with GCEH collinear, then mDHC=pi/4 and HD=b

consider similar triangles DCA and HDC,

DC/CA=HD/DC <=> x^2=a.b

Extend GCE to P such that DC = DP = x

ReplyDeleteLet < O'DC = < O'CD = < GDC = p

CD bisects < ADB and so Tr.s ADC and GDP are similar having angles 45, 45+p and 90-p

Hence x/a = b/x and the result follows

Sumith Peiris

Moratuwa

Sri Lanka