Hi

http://www.mathhelpforum.com/math-he...ture382-cg.bmp

PAQ is a double chord and BHK is a chord of the larger circle. Prove that PH is parallel to KQ.

I seriously don't know how to start this problem. Can anyone please help me?

Thanxx

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- Jan 25th 2009, 10:00 PMxwrathbringerxProving Parallelism in Circle Geometry
Hi

http://www.mathhelpforum.com/math-he...ture382-cg.bmp

PAQ is a double chord and BHK is a chord of the larger circle. Prove that PH is parallel to KQ.

I seriously don't know how to start this problem. Can anyone please help me?

Thanxx - Jan 25th 2009, 10:14 PMandreas
- Jan 25th 2009, 10:30 PMxwrathbringerx
Hmmm can I say PB and BQ are diameters of the circles?

- Jan 25th 2009, 10:47 PMandreas
- Jan 25th 2009, 11:16 PMGrandadAngles in circles
Hello xwrathbringerxYou need to join the chord AB, and then use 'Angles in the same segment are equal'. Do you know how this works? For instance, look at the angle PHB. It is

*subtended*by the arc PB at the circumference of the circle. 'Angles in the same segment' means that any other angle subtended by PB at the circumference will be equal to it. Can you see one? It's the angle PAB.

The other two facts you need concern*supplementary angles*. These are angles that add up to 180 degrees. So, for instance, angles PAB and BAQ are supplementary because they are 'Angles on a straight line'. And if you could prove that angles PHK and HKQ were supplementary, that would prove that the lines PH and KQ were parallel, because these angles are called 'Interior angles' and interior angles between parallel lines also supplementary.

So this is what you need to show:

- Angle PHB = angle PAB (as explained above)
- Angle PAQ = angle PKQ (can you explain why?)
- Angle PHK + angle HKQ = 180 degrees.

Can you fill in the gaps now?

Grandad