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June 1st, 2018, 10:54 AM   #1
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About property of triangle

Hi,

If the circle with altitude AD of triangle ABC as diameter intersects AB & AC at the point E and F, how can we prove that:

EF = $\displaystyle \frac{\Delta}{R}$

Thx.

Last edited by skipjack; June 1st, 2018 at 05:57 PM.
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June 1st, 2018, 01:55 PM   #2
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Please explain your notation thoroughly. What is $\Delta$? What is $R$?
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June 1st, 2018, 03:49 PM   #3
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Although there may be other meanings for these symbols, usually $\displaystyle \Delta =$ area of $\displaystyle \triangle ABC$ and $\displaystyle R =$ the circumradius of $\displaystyle \triangle ABC$. If this is what the OP means, then the statement is false.
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June 1st, 2018, 05:58 PM   #4
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Quote:
Originally Posted by mrtwhs View Post
Although there may be other meanings for these symbols, usually $\displaystyle \Delta =$ area of $\displaystyle \triangle ABC$ and $\displaystyle R =$ the circumradius of $\displaystyle \triangle ABC$. If this is what the OP means, then the statement is false.
Yes area for delta & circum radius is R, which are usual notations. If wrong, what EF should be equal to in terms of those two?
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June 1st, 2018, 09:30 PM   #5
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As $\displaystyle \small\triangle$ = AD$\cdot$BC/2 = (EF/sin A)BC/2 = EF(BC/sin A)/2 = EF(2R)/2, EF = $\small\triangle$/R.
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June 1st, 2018, 11:02 PM   #6
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Plz only tell me how AD = EF / sin A.

If posssible you can link me to a diagram..

Thx.
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June 2nd, 2018, 09:00 AM   #7
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Am I missing something? The area of the triangle is fixed and $\displaystyle R$ is fixed but as the diameter spins, $\displaystyle EF$ varies in length.
Attached Images
File Type: jpg Pic1.jpg (18.5 KB, 0 views)
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June 2nd, 2018, 11:35 AM   #8
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The question specifies a circle with AD as diameter. That circle is the circumcircle of triangle AEF, not triangle ABC.

As AD is a diameter of the circumcircle of triangle AEF, AD = EF/sin A is a consequence of the sine rule.

In the diagram below, the circumcircle of triangle ABC isn't shown, but its radius is R.
TriangleProperty.PNG
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June 2nd, 2018, 05:15 PM   #9
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Ahh ... I misread the problem!
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June 3rd, 2018, 05:23 AM   #10
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Quote:
Originally Posted by skipjack View Post
The question specifies a circle with AD as diameter. That circle is the circumcircle of triangle AEF, not triangle ABC.

As AD is a diameter of the circumcircle of triangle AEF, AD = EF/sin A is a consequence of the sine rule.

In the diagram below, the circumcircle of triangle ABC isn't shown, but its radius is R.
Attachment 9736
Ohk.. That's where I was confused too... It basically meant that R = cicumradius of triangle AEF (not that of ABC).

Last edited by skipjack; July 23rd, 2018 at 02:05 PM.
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