In the diagram below, A, B, C, D are four points on a circle. P is a point such taht PA and PC are tangents to the circle. D, B and P lie on a straight line. M is a point on DB such that DCM = ACB. Given that CMD = 105° and BCP = 55°, find BAC. [1]

Stuck on this question! This is not homework.

Thanks in advance.

## Circle theorem question

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### Re: Circle theorem question

tckthomas wrote:In the diagram below, A, B, C, D are four points on a circle. P is a point such taht PA and PC are tangents to the circle. D, B and P lie on a straight line. M is a point on DB such that DCM = ACB. Given that CMD = 105° and BCP = 55°, find BAC. [1]

Stuck on this question! This is not homework.

Thanks in advance.

BAC and BCP both subtend arc BC, so they are equal, so BAC = 55°. Hmmm, that seems too easy.

- dudiobugtron
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### Re: Circle theorem question

Deedlit is exactly right, but since I've never really understood what 'subtend' means in this context, I'd normally give this rule a different name:

http://www.murderousmaths.co.uk/books/bkmm7xtn.htm

http://www.murderousmaths.co.uk/books/bkmm7xtn.htm

### Re: Circle theorem question

dudiobugtron wrote:Deedlit is exactly right, but since I've never really understood what 'subtend' means in this context, I'd normally give this rule a different name:

http://www.murderousmaths.co.uk/books/bkmm7xtn.htm

No, he isn't right. BCP does not subtend BC, because those letters are in the wrong order. BPC does subtend an arc BC, but not the BC drawn in the picture, but the arc BC of the circle going through B, C and P.

You can deduce that BAC = BDC = MDC = 180-105-DCM = 75-DCM = 75-ACB = 75-ADB, but that's as far as I got.

### Re: Circle theorem question

BCP does subtend arc BC, it's just a little unusual because PC is tangent to the circle.

To see that BCP is the same angle as an angle that "normally" subtends arc BC, imagine a point Q on arc BC. BCQ will of course subtend arc BQ. Now let Q move to C. BC stays the same, and the direction of CQ approaches the direction of CP, so the angle BCQ will approach angle BCP. Meanwhile arc BQ approaches arc BC. Since angle BCQ is always 1/2 of arc BQ, BCP will be 1/2 of arc BC, same as any other angle that subtends BC.

To see that BCP is the same angle as an angle that "normally" subtends arc BC, imagine a point Q on arc BC. BCQ will of course subtend arc BQ. Now let Q move to C. BC stays the same, and the direction of CQ approaches the direction of CP, so the angle BCQ will approach angle BCP. Meanwhile arc BQ approaches arc BC. Since angle BCQ is always 1/2 of arc BQ, BCP will be 1/2 of arc BC, same as any other angle that subtends BC.

### Re: Circle theorem question

Deedlit wrote:BCP does subtend arc BC, it's just a little unusual because PC is tangent to the circle.

Ah, I see it now. Sorry, I didn't read your link (because of the horrendous colours) so didn't see what you meant. You and Deedlit are right, but I wouldn't use the term "subtend" for this situation either.

- phlip
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### Re: Circle theorem question

Another tack without using the word "subtend":

Remove D, M, P, and all their connecting lines from the diagram, they're all red herrings... just keep the ABC triangle and the tangent line at C. Then draw in O, the centre of the circle, and connect it to B and C.

The angle between BC and the tangent at C is given as 55 degrees, but I'll just call it "x" so we can be a bit more generic here. The tangent is perpendicular to the radius, and OBC is isosceles, so OCB = OBC = 90-x. Angles of a triangle add to 180, so BOC = 180 - 2*(90 - x) = 2x. Angle at centre is double angle at circumference, so BAC = BOC/2 = x.

Interestingly, since the angle-at-circumference rule holds wherever A is on the circle, then just this much gives us the answer while still having a degree of freedom in the exact layout of the points, which means we can still move around A, P, D and M to a limited extent to satisfy the CMD=105 condition... so while ignoring that given means we still can get the answer, it's still not redundant information. Just information we didn't need.

Remove D, M, P, and all their connecting lines from the diagram, they're all red herrings... just keep the ABC triangle and the tangent line at C. Then draw in O, the centre of the circle, and connect it to B and C.

The angle between BC and the tangent at C is given as 55 degrees, but I'll just call it "x" so we can be a bit more generic here. The tangent is perpendicular to the radius, and OBC is isosceles, so OCB = OBC = 90-x. Angles of a triangle add to 180, so BOC = 180 - 2*(90 - x) = 2x. Angle at centre is double angle at circumference, so BAC = BOC/2 = x.

Interestingly, since the angle-at-circumference rule holds wherever A is on the circle, then just this much gives us the answer while still having a degree of freedom in the exact layout of the points, which means we can still move around A, P, D and M to a limited extent to satisfy the CMD=105 condition... so while ignoring that given means we still can get the answer, it's still not redundant information. Just information we didn't need.

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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: Circle theorem question

I posted an answer but this was incorrect sorry.

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