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 October 28th, 2008, 02:42 PM #1 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Sum of distances from sides of triangle equals height Let P be any point inside an equilateral triangle. Show that the sum of the distances of P from the three sides is always equal to the height of the triangle. (The distances are the shortest possible, i.e., they form right angles with the sides of the triangle.)
 October 28th, 2008, 07:35 PM #2 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Sum of distances from sides of triangle equals height Draw the perpendiculars and also join the point to the vertices. Let the side of the isosceles triangle be s, and the perpendiculars a, b, c. Now, figure the areas of the three internal triangles for which the perpendiculars are their heights and form their sum. Also, let the height of the isosceles triangle be "h", and figure its area. There you have it.
 October 28th, 2008, 08:54 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Sum of distances from sides of triangle equals height Thanks.
 October 28th, 2008, 10:38 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Sum of distances from sides of triangle equals height So $A= \frac{1}{2}sa+\frac{1}{2}sb+\frac{1}{2}sc=\frac{1} {2}s(a+b+c)\mbox{ so }\frac{A}{\frac{1}{2}s} = h = a+b+c$

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