geometry question (1 Viewer)

[5647382910]

ABCD is a trapezium with AB parallel to DC. the diagonals intersect at O. the line through O parallel to AB cuts AD and BC at P and Q respectively. prove that PO = QO.

thanks =]

 

[Timothy.Siu]

u can use similar triangles,

OQ/DC=BO/BD

PO/DC=AO/AC

DC=PO.AC/AO
DC=OQ.BD/BO

OQ.BD/BO=PO.AC/AO

OD/BO=OC/AO (similar triangles)

1+OD/BO=1+OC/AO

(BO+OD)/BO=(AO+OC)/AO

BD/BO=AC/AO

therefore, OQ=PO

 

[5647382910]

u can use similar triangles,

OQ/DC=BO/BD

PO/DC=AO/AC

DC=PO.AC/AO
DC=OQ.BD/BO

OQ.BD/BO=PO.AC/AO

OD/BO=OC/AO (similar triangles)

1+OD/BO=1+OC/AO

(BO+OD)/BO=(AO+OC)/AO

BD/BO=AC/AO

therefore, OQ=PO

ok i see, thanks heaps