Produce C E until E D equals C E, and join F D. The two triangles F E C, F E D, have the side C E of the one equal to the side E D of the other, the side F E common, and the included angle F E C of the one equal to the included angle F E D of the other, they are therefore equal, and the side C F equals the side F D. But the straight line C D is the shortest distance between the two points C D; therefore it is shorter than the broken line C F D. Then C E, the half of C D, is shorter than C F, the half C F D. And, as C F is any line other than a perpendicular, the perpendicular C E is the shortest line that can be drawn from C to A B. [Illustration] PROPOSITION XVII. THEOREM. DEMONSTRATION. We wish to prove that _A tangent to a circumference is perpendicular to a radius at the point of contact._ Let the straight line A B be tangent at the point D to the circumference of the circle whose centre is C. Join the centre C with the point of contact D, the tangent will be perpendicular to the radius C D. For draw any other line from the centre to the tangent, as C F. As the point D is the only one in which the tangent touches the circumference, any other point, as F, must be without the circumference. Then the line C F, reaching _beyond_ the circumference, must be longer than the radius C D, which would reach only to it; therefore C D is shorter than any other line which can be drawn from the point C to the straight line A B; therefore it is perpendicular to it. PROPOSITION XVIII. THEOREM. DEMONSTRATION. [Illustration] We wish to prove that _In any isosceles triangle, the angles opposite the equal sides are equal._ Let the triangle A B C be isosceles, having the side A B equal to the side A C; then will the angle B, opposite the side A C, be equal to the angle C, opposite the equal side A B. For draw the line A D so as to divide the angle A into two equal parts, and let it be long enough to divide the side B C at some point as D. Now the two triangles A D B, A D C, have the side A B of the one equal to the side A C of the other, the side A D common to both, and the included angle B A D of the one equal to the included angle C A D of the other; therefore the two triangles are equal in all respects, and the angle B, opposite the side A C, is equal to the angle C, opposite the side A B. [Illustration] PROPOSITION XIX. THEOREM. DEMONSTRATION. We wish to prove that, _If two triangles have the three sides of the one equal to the three sides of the other, each to each, they are equal in all their parts._