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April 9th, 2018, 12:12 AM   #1
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How angle ACB is 90?

In circle ABC where O is the center, there is a triangle ABC. but how angle ACB is 90?

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April 9th, 2018, 12:52 AM   #2
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That result is known as Thales's theorem.
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April 9th, 2018, 04:49 AM   #3
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A more general theorem is that if an angle is inscribed in a circle (all three vertices on the circle) then the angle measure is half the measure of the arc it cuts off. Since AB goes through the center, O, it is a diameter. What is the measure of the arc AB? So what is the angle ACB?
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April 9th, 2018, 04:50 AM   #4
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The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at the circumference of that circle.



The angle AOB is twice the angle ACB.

********************
In our case, the angle AOB = 180 degrees, so the angle ACB must be half of the angle AOB = (1/2)*(180) = 90 degrees.
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April 9th, 2018, 07:51 AM   #5
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As triangles AOC and BOC are isosceles, $\small\angle$ACB = $\small\angle$ACO + $\small\angle$BCO = $\small\angle$CAO + $\small\angle$CBO.
Hence $\small\angle$ACB is half the sum of the internal angles of triangle ABC.
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April 9th, 2018, 08:54 AM   #6
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Quote:
Originally Posted by Hussain2629 View Post
......
Please try to minimize the image size from next time.
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April 9th, 2018, 11:10 PM   #7
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Quote:
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Please try to minimize the image size from next time.
Sorry for that dear.. I will minimize next time.... I hope my answer was clear for you.
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April 16th, 2018, 07:48 AM   #8
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Is it obvious to me alone?
Using Thales's theorem is all easy. Usually if I do not understand something, I'll first read all the relevant information, if I have not found my answer, then I ask other sources starting from mathforums ending with help sites in homework. Thanks to this, I now feel much more strongly in various fields of activity. Learn while young.

Last edited by skipjack; April 16th, 2018 at 08:12 AM.
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April 16th, 2018, 09:01 AM   #9
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Scotty, can we buy you a one-way ticket to another site?
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April 16th, 2018, 10:45 AM   #10
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Behave, Denis!
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