In a circle, a diameter bisects the angle formed by two intersecting chords.
Prove that the chords are equal
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Kevin
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I've tried drawing two chords and the diameter bisecting the angle at the intersection, then trying to say that the triangles at the intersection and the circumference of the circle are equal, but I can't seem to convince myself of that using Euclids propositions. I've also tried to show that the diameter is the perpendicular bisector to the line segment connecting the two points, but I am also having trouble convincing myself of that – Kevin Mar 10 '16 at 09:15
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draw a picture please – Dr. Sonnhard Graubner Mar 10 '16 at 09:15
2 Answers
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Triangles IEB IEC are congruent (sides and included angle common)
The three lines are $ concurrent $ at E. $Three $ vertically opposite angles are same. Mark them separately as $ p,q,r $. Choose from among them conveniently .
BE = EC
Triangles AEI DEI are congruent (sides and included angle common)
DE = AE
Total chord length is same;
AE + EC = DE + EB.
Narasimham
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I've been stuck on showing that the included angles are actually equal. Could you specify how you found this – Kevin Mar 10 '16 at 10:00
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Yes but you're also saying that two of the vertically opposite angles are equal. I'm just not sure how you can prove that the diameter bisects $I$ as well – Kevin Mar 10 '16 at 10:08
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You meant $ I $bisects diameter. Where did we assume in our proof that $I $ is the center of circle or diameter midpoint? – Narasimham Mar 10 '16 at 10:12
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Please read the steps carefully. Forget $I$. The mod may ask us to go to chat – Narasimham Mar 10 '16 at 10:21
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Angle AEB= angle CED (vertical angles are equal)
Arc AC = arc BD (arc addition postulate)
AC=BD (equal arcs have equal chords)
georgina
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