Circle Geometry (1 Viewer)

withoutaface

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Through extending some lines to H (midpoint of XY) and the midpoints of AB and DC we can create some right angles and apply pythagoras' theorem to get:
HY^2=r^2-OH^2
MD^2=r^2-OM^2

and since HY=XY/2 and MD=CD/2, then if XY>CD then r^2-OH^2>r^2-OM^2
OM^2>OH^2

This is evidently true if we consider the right angled triangle OMH, because OM is the hypotenuse, and is hence greater than OH, so XY>CD.

A similar argument can be applied for AB.
 

Mellonie

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confused with ur way
i follow ur argument but i seem to be lost on the last line

i just dont c how by proving OM^2> OH^2

proves XY is greater then CD

I am pretty sure they want us to use congruency here to prove LH= MH
and then prove LH= LA= BL = CM = MD
so then we can say XL + MY r the remaing line of XY

HEnce XY is larger

but hey u hav helped too se som things i was mising
 

withoutaface

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Basically we know OM^2>OH^2 because it's the hypotenuse of a right angled triangle. If we work backwards:
-OH^2>-OM^2
r^2-OH^2>r^2-OM^2
(XY/2)^2 > (CD/2)^2 (pythagoras)
XY^2>CD^2
XY>CD

And we apply the same kind of argument to AB.
 

Mellonie

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withoutaface said:
Basically we know OM^2>OH^2 because it's the hypotenuse of a right angled triangle. If we work backwards:
-OH^2>-OM^2
r^2-OH^2>r^2-OM^2
(XY/2)^2 > (CD/2)^2 (pythagoras)
XY^2>CD^2
XY>CD

And we apply the same kind of argument to AB.


BUt my point is thAT I THINK UR MENT TO PROVE BY CONGRUENCY..
CAN U SEE ANY OTHER WAY TO PROVE
 

who_loves_maths

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Originally Posted by Mellonie
BUt my point is thAT I THINK UR MENT TO PROVE BY CONGRUENCY..
CAN U SEE ANY OTHER WAY TO PROVE
i THINK that what you THINK is wrong. unless the question specifically asks for proof by congruency, which it does not since otherwise you wouldn't be THINKING that it does, then what withoutaface has done here is perfectly legitimate... it's also probably the fastest method to doing this question.

stop making life difficult for yourself... plus, in the new thread that you started, which has since been closed, why didn't you specify the "needs to be proven via congruency" 'requirement'???
 

Mellonie

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...

i am not saying its not legitimate.. i just dont c it...

it is not my fault i have tried but it doesnt make sence to me. THe section i got the question from all required congruency.. ike it was a prove by congruency section so i thuoght maybe theirs a way to do that
 

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