A straight line a is drawn through the middle of the segment AB. Perpendiculars AC and BD

A straight line a is drawn through the middle of the segment AB. Perpendiculars AC and BD are drawn from points A and B to line a. Prove that AC = BD

Let us denote the middle of the segment AB by the letter O.

If you release the perpendiculars to the straight line a, AC and BD from points A and B, you get two right-angled triangles AOC and BOD.

These triangles will be equal, since AO = OB (O-midpoint of the segment AB). Angle AOC = angle DOB (vertical angles are equal).

So right-angled triangles are equal in hypotenuse and acute angle.

From the equality of the triangles it follows that AC = BD.