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Please help with these Geometry question (1 Viewer)

a1079atw

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Hi guys,

I'm having trouble with these 3 questions... can anyone please help me with them?
Screen Shot 2013-06-30 at 8.54.50 PM.jpg (for question 2, only (a) thanks)
Screen Shot 2013-06-30 at 8.55.05 PM.jpg

Thanks in advance!
 

Drongoski

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Q3

Let the 2 circles intersect at A & B, with A on the line PQ, and B on RS

Let angle APR = @

.: angle PRB = 180 - @ ( APR & PRB are co-interior angles and PQ//RS)

.: angle PAB = @ (ABRP is cyclic quad, PRB + PAB = 180)

.: angle QSB = angle PAB = @ (ABSQ a cyclic quad; ext angle of a cyclic quad = interior opp angle)

.: angle PRS + angle QSR = 180 (i.e. the 2 angles are supplementary)

But PRS & QSR are co-interior angles

.: PR//QS

.: PQSR is a paralellogram (opp sides are parallel)

.: PQ = RS (opp sides of //gram)


QED
 
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Drongoski

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Q2(a)

This one is straightforward. Since angle HEC + angle HDC = 180 (these 2 angles, 90 deg each, are supplementary)

But these 2 angles are opposite angles of quadrilateral HDCE.

.: HDCE is cyclic.
 
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a1079atw

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Thanks a lot!! For question 2, I thought you have to prove that both pairs of opposite angles add up to 180 lol, and now I realised it's a quadrilateral so if one pair adds up to 180 then the other pair must be so too.

Anyone knows how to do question 1?
 

a1079atw

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Thanks, I got (a) now, but still have no idea about (b). I tried to prove that angle BAX= angle CAX= x, and that angle ACX= 2*(x) but I don't know how to prove that latter part, angle ACX= 2*(x).
 

nightowl

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Thanks, I got (a) now, but still have no idea about (b). I tried to prove that angle BAX= angle CAX= x, and that angle ACX= 2*(x) but I don't know how to prove that latter part, angle ACX= 2*(x).
Are you allowed to refer to Ptolemy's theorem without proof? If so question 1b becomes trivial.
 

nightowl

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:)

In that case let angle BAX= and we easily get the following angles.

XAC=

ACX=

Without loss of generality let AB=AC=BC=1.

Then, using the sine rule







Trig expand and simplify and it's done.
 

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