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 June 1st, 2018, 10:54 AM #1 Senior Member     Joined: Jan 2012 Posts: 123 Thanks: 2 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.
 June 1st, 2018, 01:55 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond Please explain your notation thoroughly. What is $\Delta$? What is $R$?
 June 1st, 2018, 03:49 PM #3 Senior Member     Joined: Feb 2010 Posts: 697 Thanks: 135 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.
June 1st, 2018, 05:58 PM   #4
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Quote:
 Originally Posted by mrtwhs 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?

 June 1st, 2018, 09:30 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,003 Thanks: 1862 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.
 June 1st, 2018, 11:02 PM #6 Senior Member     Joined: Jan 2012 Posts: 123 Thanks: 2 Plz only tell me how AD = EF / sin A. If posssible you can link me to a diagram.. Thx.
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
 Pic1.jpg (18.5 KB, 0 views)

 June 2nd, 2018, 11:35 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,003 Thanks: 1862 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
 June 2nd, 2018, 05:15 PM #9 Senior Member     Joined: Feb 2010 Posts: 697 Thanks: 135 Ahh ... I misread the problem!
June 3rd, 2018, 05:23 AM   #10
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Quote:
 Originally Posted by skipjack 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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