Chord parallel to AB and 2CD=AB

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    Chord Parallel
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Discussion Overview

The discussion revolves around a geometric problem involving a circle, a diameter AB, and a chord CD that is parallel to AB with the condition that 2CD=AB. Participants explore methods to prove that AE=2AB, discussing various geometric properties and relationships.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the problem and considers using the relationship AE*CE=BE² but finds it insufficient.
  • Another participant suggests that if A and C are on the same side of the chords, then triangle OAC is equilateral, leading to the conclusion that angle BAE is 60 degrees, which implies AE=d and AB/AE = cos60 = 1/2.
  • A different participant proposes constructing a trapezoid with CD and AB, asserting that triangle BAE is a right triangle due to the angle between the radius and tangent being 90 degrees. They provide a series of computations leading to the conclusion AE=2AB.
  • One participant mentions having an illustration of the triangle and trapezoid but cannot upload it due to scanner issues.
  • A participant acknowledges their progress in forming the equation and recognizes that it required manipulations and rearrangement, confirming their understanding of the equilateral triangle concept.

Areas of Agreement / Disagreement

Participants express varying approaches to the problem, with some agreeing on the equilateral triangle concept while others focus on different geometric constructions. The discussion remains unresolved as multiple methods and interpretations are presented without consensus.

Contextual Notes

Some assumptions about the configuration of points A, B, C, and D are not explicitly stated, and the dependence on geometric properties such as the nature of angles and triangle relationships is implied but not fully explored.

himanshu121
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AB is the diameter of a circle CD is a chord parallel to AB and 2CD=AB. The tangent at B meets the line AC produced at E. Prove that AE=2AB.

I'm finding no way to solve this ?

Still i thought of applying AE*CE=BE2 but that is not enough to slove the pro ?

Any other Hint to solve
 
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So A and C are on the same end of the chords?

Let O be the centre of the circle

the triangle OAC is equilateral, because the triangle OCD is and everything is symmetric.. well, convince yourself somehow

so angle BAE is 60,

so AB/AE = cos60 = 1/2; AE=d. QED

Opposite ends of te chords gives you...?
 
i think you should bulid a trapozoid with CD and AB (with two sides equals to each other and to the small base-CD) and to see that we have THE TRAINGLE BAE as a right triangle because the angle between radius and a tangent is 90.
here are my computations that brought to the answer:
BE^2=CE*AE
AB^2+BE^2=AE^2 (PYTHAGORAS THEOREM)
AB^2+CE*AE=AE^2
AB^2=AE(AE-CE)=AE*CE
CE=CD=AB/2
AB^2/CE=AE
AB^2/(AB/2)=AE
2AB=AE

so you were half right himanshu :wink:
 
i have an illustration of the triangle and trapaoid the problem is my scanner doesn't work so i can't upload it here, sorry.
 
Ya i got it
I was half way write upto formation of equation and it just required manipulations & rearrangement

Ya i convinced myself for equilateral triangle
Thnxs
 

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