3u circle geometry (1 Viewer)

[mack5]

this is a hard Q, can anyone help? -- AB, CD adn XY are chords in a circle with centre O. XY cuts AB and CD in L and M, which are the midpoints of AB and CD. Prove thta XY is greater than either AB or CD. ---

 

[rithvikt]

this is a hard Q, can anyone help? -- AB, CD adn XY are chords in a circle with centre O. XY cuts AB and CD in L and M, which are the midpoints of AB and CD. Prove thta XY is greater than either AB or CD. ---

if u had put up a picture of the prob that would have been helpful...

 

[m_isk]

which book is this from? pls provide page number

 

[HellVeN]

I drew the diagram.

I can't be fucked solving it though

 

[Trebla]

XL x LY = BL²
XM x MY = DM²

Lol, thats all I came up with. There's a start. I think you might need to use the isoscles triangle formed from the centre to chords AB and DC since AB _|_ LO and DC _|_ OM. Have fun solving the problem! lol!

 

[withoutaface]

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.

 

[m_isk]

You must spread some Reputation around before giving it to withoutaface again.