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by Matt@VeritasPrep » Thu Jun 23, 2016 3:43 pm
Any time you have an inscribed angle, you can draw a central angle to the same two points such that the central angle is DOUBLE the inscribed angle.

So if <AED = 75°, <AOD = 150° (where O is the center of the circle).

From there, <AOB, <BOC, and <COD sum up to 150° and are each the same, so

<AOB = <BOC = <COD = 50°.

∆BOC is isosceles (since two of the sides are the radius of the circle), so its angles are 50°, 65°, and 65°. x is two of the base angles, or 2*65° => 130°.
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by evs.teja » Wed Jun 29, 2016 12:44 pm
Matt@VeritasPrep wrote:Any time you have an inscribed angle, you can draw a central angle to the same two points such that the central angle is DOUBLE the inscribed angle.

So if <AED = 75°, <AOD = 150° (where O is the center of the circle).

From there, <AOB, <BOC, <COD sum up to 150° are each the same, so

<AOB = <BOC = <COD = 50°.


Dear Matt,
Just a small clarification,
All these angles are equal coz all of them (OAB,OBC,OCD) are congruent to each other via SSS congruence theorem. right ?

Regards
Teja
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by OptimusPrep » Wed Jun 29, 2016 8:12 pm
The angles are same because AB = BC = CA
Hence the angles subtended by these chords on the centre will also be same, as will be the arcs.

Does this help?
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by evs.teja » Thu Jun 30, 2016 2:42 am
OptimusPrep wrote:The angles are same because AB = BC = CA
Hence the angles subtended by these chords on the centre will also be same, as will be the arcs.

Does this help?
Thank you sir ....yes sir it completely made sense....completely forgot this rule.

Regards
Teja
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