If diameter AB of a circle bisects a chord PQ, and AQ is parallel to PB, show that PQ is equal to AB in length
2007-03-06 01:25:54 · 12 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Call the point where PQ and AB intersect "C".
If AQ is parallel to PB then
angle BPQ is congruent to angle AQP
AND
angle PBA is congruent to angle QAB
because they are alternate interior angles.
We already know that PC = CQ because it was said that AB bisects PQ.
So, triangle AQC and triangle BPC are congruent through Angle Angle Side (AAS).
Then PC = CQ and AC = CB because corresponding parts of congruent triangles are congruent.
If AC = CB, then C would be the midpoint of AB and AB is the diameter, then C must be the center of the circle.
Also, C is the midpoint of PQ, so PQ must also go through the center of the circle. So, PQ is also a diameter. All diameters of a cirlce are congruent, so PQ = AB
2007-03-06 01:47:00 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
in order for PB and AQ to be parallel, AB and PQ have to meet at the center of the circle. If any chord passes through the radius, it is considered a diameter. There are some geometric proofs to this, but i can't figure them out.
The best way i could show it would be to draw to parallel chords, AQ and PB.. then draw your diameters AB and PQ.. you'll notice that no matter where the two parallel lines are drawn, the two chords both meet at the center.
2007-03-06 09:47:09 · answer #2 · answered by Amanda 2 · 0⤊ 0⤋
I don't think that's necessarily true. I just drew a picture of a circle, with diameter AB, which in turn bisects chord PQ. For AB to equal PQ, PQ would have to be another diameter of the circle. But you can draw this where P and Q are not points on the circle, and all of your criteria are met.
Sorry I can't be of more help...
2007-03-06 09:36:09 · answer #3 · answered by Dave 6 · 0⤊ 1⤋
The only way it is possible is if PQ=AB, i.e. the chord is the diameter.
2007-03-06 09:50:29 · answer #4 · answered by verbalise 4 · 0⤊ 1⤋
2 points
2007-03-06 09:51:43 · answer #5 · answered by Anonymous · 0⤊ 1⤋
hey,you know that PO*OQ=AO*OB (EQ.1)
now triang. POB is similar to triang. AOQ(SINCE ang.OQA=ang.OPB,ang.POB=AOQ & ang.OBP=ang.OAQ i.e by AAA SIMILARITY)
PO/AO=OB/OQ=PB/AB , by EQ.1
PB=AB.
NOW by ASA congurency in those same triang. triang.POB in congurent to triang. AOQ.
THUS AQ=PB.
2007-03-06 09:48:09 · answer #6 · answered by SS 2 · 0⤊ 0⤋
He shoots, He scores! Two points!
2007-03-06 09:34:43 · answer #7 · answered by helloeveryone 3 · 1⤊ 1⤋
I can't draw on here therefore I can't show you the answer. sorry.
2007-03-06 09:29:49 · answer #8 · answered by Anonymous · 0⤊ 1⤋
it looks like a crosshair?
2007-03-06 09:38:01 · answer #9 · answered by thelittlelim 2 · 0⤊ 1⤋
No.. lol
2007-03-06 09:33:08 · answer #10 · answered by 04/12/2008 :) 6 · 0⤊ 1⤋