How is the angle bisector created in the diagram?
The given explanation is: Let O be the center of the smaller circle. The lines CO and DA are parallel (both perpendicular to DB). Consequently, the angles DAB and COB are equal.
Now in the small circle the angle CAO divides a chord, the angle of which at the corresponding center is called COB.
Consequently: CAO = 1/2 COB = 1/2 DAB.
Thus CAO = DAC: AC is then the angle bisector of DAB.
It's precisely the "consequently" that I don't understand. Does anyone have a plausible explanation?
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Triangle OAC is equal
Angle CAO = angle OCA
Angle COB = external angle to triangle OAC
Angle COB = angle DAB = angle CAO + angle OCA
Wait…
Okay, get it.
Thank you
Hello,
From C, take a tendon of the small circle to the point at which the small circle and the diameter of the large circle AB intersect.
Name this intersection E.
Then the peripheral angle set is valid in the small circle.
The center angle EOC is twice as large as the periphery angle EAC over the same chord CE.
Since EOC=BAD, the assertion is correct.
Best regards,
Willy
The point of contact is A, right?
The other intersection. The diameter cuts the small circle not only at one point.
Okay.
I hope you don’t mind if I give you the most helpful.