## 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, AY^{2} = 2^{2} + 1^{2} AY = 5.

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

Therefore DX/2=1/5 DX = 2/5.

Hence triangle DXC is isosceles (CD=CY=1), but not equilateral (DX=2/5).

By labelling D as the origin,

- find the equation of the line passing through A and X.
- 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.
- find the point of intersection of the two lines, X.
- use this information to solve the problem.

Problem ID: 134 (Nov 2003) Difficulty: 3 Star