Circle question (1 Viewer)

[namburger]

http://i237.photobucket.com/albums/ff3/Namburger/circlegeoquestion.jpg

Help
Thanks in advance

 

[independantz]

i) <ADC=<ACB=90( the angle between a tangent and radius at the point of contact is 90)

therefore, BCAD is a cyclic quad( opp angles are supplementary)

ii)given: the circles have equal radii

therefore, AE=EB, thus E is the centre of AB
From i), <ACB= 90
thus, AB is the diameter of circle BCAD
Therfore, E is the centre of the circle

iii)let AC=r,
thus, AB=2r
sin<CBA= r/2r=1/2
thus, <CBA=30

similarly for the other angle...

iv)from iii) <DBA=30, thus <DAB= 60 ( angle sum of a triangle
thus, <DCB= 60 ( angles in the same segment are equal when standing on the same arc)
therfore <DCB=<CBD= 60
thus, <CDB=60 ( angle sum of a triangle


Hence, triangle BCD is equilateral, ( all angles equal 60)

Here's the attachment if you cant see the text lol

 

[independantz]

^ WTf is it just me or is half the text up there missing?

when i click edit i can see the text, however half of it is missing in the post lol?

 

[qqmore]

(ii) < ACB = < ADB = 90 degrees (tangents perpendicular to radius)
<ACB + <ADB = 180 degrees
BCAD is cyclic (opposite angles added up to 180 in a cyclic quadrilaterial)

(iii)
Since< ACB = 90
and < ADB = 90

Thus, AB must be diameter (< in semi circle = 90)

But, AE = EB (=equal radius of equal circles)

Thus E is the centre of the circle passing through B,C,A,D

(iv)??




(v) from iv) < CBD = 60

But CB = DB (tangents from external point are =)

Thus, < DCB = < CDB

< DCB + < CDB + 60 = 180 (< sum of triangle)
2< DCB = 120
<DCB = 60
and it follows < CDB = 60 degrees

Therefore triangle BCD is equaliteral, as all angles are equal.

 

[namburger]

SHIT, didnt see the equal radii part. Thanks

dw independent, i know how to do it