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Perpendicular Construction

Problem

ABCD is a unit square, M is the midpoint of BC, and DX is perpendicular to AM.


Prove that triangle DXC is isosceles but not equilateral.


Solution

Consider the following diagram.


It is possible to produce a semi-circle, with centre at C and radius CD. As X lies on the semi-circle, triangle DXY will be right angle, and length CX will be equal to the radius, 1 unit. Hence lengths CD and CX are equal and we have proved that the triangle is, at least, isosceles.

Using the Pythagorean Theorem, AY2 = 22 + 12 implies AY = radical5.

By similar triangles, DX/DY = AD/AY.

Therefore DX/2=1/radical5 implies DX = 2/radical5.

Hence triangle DXC is isosceles (CD=CY=1), but not equilateral (DX=2/radical5).

By labelling D as the origin,

  1. find the equation of the line passing through A and X.
  2. use the result that the product of gradients is -1 iff two lines are perpendicular to find the equation of the line through D and X.
  3. find the point of intersection of the two lines, X.
  4. use this information to solve the problem.
Problem ID: 134 (Nov 2003)     Difficulty: 3 Star

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