Saturday, May 8, 2010

Problem 448: Tangent Circles, Diameter, Perpendicular, Chord

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

Complete Problem 448
Level: High School, SAT Prep, College geometry

1. DCE = DAB = 1/2 DO'E = DOB ( DO'E, DOB isoc, ODB comm )
DEC = DCA ( have same arc DC )
=>
DC common for ADC, DCE (2)
(1) & (2)
=>
AD = DC = DE = x (3)

▲DGE ,
GDE = 90 - ADG = 90 - B
GED = 90 - B (ADC isoceles => DG bisector )
=>
DGE isoceles =>
=>
x/a = b/x

x² = ab
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1. Tr.s ADC and DCE are not necessarily congruent. This will be so only in the special case of each triangle being isoceles.
Could u elaborate?

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

3. Extend GCE to P such that DC = DP = x

Let < 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