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In a circle, AB be the diameter.. X is an external point. Using straight edge construct a perpendicular to AB from X
If X is inside the circle then how can this be done

Discussion:

Teacher: What fascinates me about CMI problems is that they are at once fundamental and beautiful in nature. This problem which concerns straight edge construction breaths on a very simple geometric fact: you cannot create a right angle with straight edge until it is already there. Try using this hint.

Student: What is a straight edge? Is it a ruler?

Teacher: A straight is a ruler without markings. When you are constructing with straight edge, you are basically allowed to join given points and nothing else. Although another thing that you can do is select new points from the plane.

Student: Since I cannot think of doing anything in particular, let me join XA and XB.

Teacher: That is indeed the first step. Can you think of finding a right angle anywhere? Remember AB is a diameter.

Student: Oh! I see this. Suppose XA and XB intersects the circle at P and Q. $\angle APB$ and $\angle BQA$ are right angles (as AB is the diameter and these are angles subtended by diameter at the circumference, it is common geometric fact that they must be right angles).

Teacher: Excellent. Suppose AP and BQ intersect at H. Now finish this of.

Student: Sure. Here H is the orthocenter of triangle AXB. Sure the line XH (when produced) will be perpendicular to AB as well (as altitudes of a triangle are concurrent).

The second part of the problem can be done similarly. Here is a diagram that will be ‘proof without words’.

Teacher: Very well. CMI 2014 also had a problem on straight edge construction. In fact there is this famed theorem (due to Mohr- Mascheroni) that every construction that can be done by compass and straight edge can be done by a compass alone and (more importantly), any construction doable by a compass can be done by straight provided you have a circle given.