Circle Geometry Question (2 Viewers)

[nrlwinner]

Hi. This is my first post here although I've been reading many posts before. I need some help on this question. I mainly need a diagram because I don't get what the question is trying depict. If you can, leave the answer and working out out, but include a diagram.

Cheers.

22. Two points A and B are taken on a circle and C is the other end of the diameter through A. If AE is the perpendicular from A on to the tangent at B, prove that AB bisects angle CAE.

 

[jet]

CAE = 90° (given)
Therefore AE is the tangent to the circle at A
Therefore angle EAB = angle BCA (alternate segment theorem)
Now, angle ABC = 90° (the angle in a semicircle is a right angle, AC is a diameter)
Therefore angle BAC = 90 - BCA = 90 - EAB

But, since EAC = 90° angle BAC already equals 90 - EAB
Hence BAC = EAB
Therefore AB bisects angle CAE.

 

[EvoRevolution]

where did u draw the diagram was tryin to find somethin online but nothin

 

[jet]

in Pages '09

 

[Drongoski]

Here's soln. But I don't know how to get diagram up.

Draw a circle centre o. Pick two points a & b on circle. Draw diameter thru A & O to meet circle at C. Draw tangent line at B and drop a perpendicular from A to this tangent at E. Draw lines AB and BC. Let F be any point on tangent opposite side of B from E .

Now angle OBE is a right angle, Therefore AE//OB.
Now angle CBF = angle CAB = @ say ( angle in alt segment)
Since triangle AOB is isosceles, angle OBA = angle BAO = @
Angle EAB = angle OBA = @ (alt angles , AE//OB)

Therefore angle EAB = angle BAC

Hence AB bisects angle CAE


QED

jetblack2007: very nice diagram; wish I knew how to do it.

Edit

Oh what an embarassingly long-winded solution. Here's revised version; forget F)

angle BAO = angle ABO (ABO isosceles)
But angle EAB = angle ABO (alternate angles, AE//OB)
Therefore angle OAB = angle EAB
Hence AB bisects angle CAE

 

[EvoRevolution]

for mac

 

[jet]

yup.

 

[EvoRevolution]

awesome

 

[Drongoski]

jetblack2007:

have I read the question incorrectly? Isn't AE to be perpendicular to the tangent at B; i.e. AEB is 90 deg. ??

 

[GUSSSSSSSSSSSSS]

jetblack2007:

have I read the question incorrectly? Isn't AE to be perpendicular to the tangent at B; i.e. AEB is 90 deg. ??

That's the impression i got when reading the question
and did it that way u did in the "edit" part of ur solution

..

 

[jet]

I got the impression that it was perpendicular to A, because it said the perpendicualr from A to B, as in the perpendicular is originating at A.
Though it is weird that both solutions worked. So I am really unsure now.

 

[EvoRevolution]

yea i though that aswell

 

[GUSSSSSSSSSSSSS]

I got the impression that it was perpendicular to A, because it said the perpendicualr from A to B, as in the perpendicular is originating at A.
Though it is weird that both solutions worked. So I am really unsure now.

lol i think it's STILL just as ambiguous lol

but yes, very odd that both ways worked out in the end hmmm......

 

[Drongoski]

No. Re perpendicular, there is no ambiguity at all.

 

[GUSSSSSSSSSSSSS]

No. Re perpendicular, there is no ambiguity at all.

so u wud agree with jetblack's diagram??

 

[Drongoski]

so u wud agree with jetblack's diagram??

Not at all! His is for a different problem.

 

[Timothy.Siu]

yeah i thought it was what drogonski said

 

[GUSSSSSSSSSSSSS]

kk im confused

i initially thought the same as drongoski

but then his strong reply made me think he agreed with jetblack so my head start turning in that direction

then drongoski say he DUN agree with jetblack
and tim siu agrees

but it was odd how in both instances it worked out perfectly...hmmmmmmm

 

[jet]

Well, it doesn't even matter really... the OP has a choice of both.
Problem solved

 

[lyounamu]

Well, it doesn't even matter really... the OP has a choice of both.
Problem solved

golden answer

+reppped