Math Is Fun Forum

naturewild

Member

Registered: 2005-12-04

Posts: 30

Proving with vectors

1) Prove that the median to the base of an isosceles triangle is perpendicular to the base.

2) In triangle ABC, the points D, E, and F are the midpoints of sides BC, CA, and AB, respectively. The perpendicular at E to AC meet the perpendicular at F to AB at the point Q.

a. Prove that AB dot (QD - ½AC) = 0

b. Prove that AC dot (QD - ½AB) = 0

c. Use parts a and b to prove that CB dot QD = 0

d. Explain why these results prove that the perpendicular bisector of the sides of a triangle meet at common point.

Thank you in advance!!

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Proving with vectors

1.  In an isosceles triangle, the two sides are the same length and so are the angles.  From this, you can conclude that the two triangles (one on each side of the median) are similar by SSA.  Then you can conclude the base angles of the median are equal to one another.  So they must be 90.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

naturewild

Member

Registered: 2005-12-04

Posts: 30

Re: Proving with vectors

Thanks Ricky... but I need to prove it with vectors method. Either position vector or point-to-point vector.

Troublesome I know

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Proving with vectors

Sorry, it would help if I read the subject line once in a while

This isn't a problem, as long as you set it up right.  We need to set up our triangle using vectors in such a way that we don't lose any generalities.  That means a lot of variables.

First thing we need is our vertex, (a, b).  Now we need to go down and to the right.  So let epsilon and delta > 0 (e and d from here on).  Also, let the left point be (a - e, b + d) and the right point be a + e, b + d)

Then our two vectors spanning the left and right sides of the triangle are:

Left: (a - e, b + d) - (a, b) = (-e, d)
Right: (a + e, b + d) - (a, b) = (e, d)

We want a vector which splits the other two in half (our median), so we add the left and the right together:

median = (-e, d) + (e, d) = (0, 2d)

Now we want to find the vector spanning the bottom of the triangle, which is the vector from the two points where our left and right vectors ended:

Bottom: (a + e, b + d) - (a - e, b + d) = (2e, 0)

All that's left is to show that (2e, 0) and (0, 2d) are perpendicular, which can be done by taking the dot product.

Last edited by Ricky (2006-04-30 11:53:42)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."