Prove existence and uniqueness of a perpendicular to a line

Prove existence and .uniqueness of a perpendicular to a line at a point on the line.

Solution

P

| \\
|. \\
|. . \\
|. . . \\
|____\\___________d
A. . . B

Let P be your point exterior of the line d, and PA be perpendicular on d.

The proof you ask for is easier done by contradiction: assume there exists another perpendicular from P to d. Let this other perpendicular be PB.

(1) If PA and PB are different lines having only P in common then A is different from B ==> PAB is a triangle ==> the measures of the angles P, A and B are all greater than 0 degress and less than 180 degrees AND the sum of the measures of the three angles is exactly 180 degrees.
(2) PA_|_d ==> measure of angle PAB is 90 degrees
(3) PB_|_d ==> measure of angle PBA is 90 degrees

Putting together (1), (2) and (3) we conclude that:
measure_of_PAB + measure_of_PBA + measure_of_APB = 180 degrees
90 degrees + 90 degrees + measure_of_APB = 180 degrees
180 degrees + measure_of_APB = 180 degrees
measure_of_APB = 180 degrees - 180 degrees
measure_of_APB = 0 degrees ==> PA and PB are the same line. This is a contradiction with the fact in (1), that is PA and PB are different lines having only P ==> there does not exist another perpendicular from P to the line d.