The perpendicular line from a point to a straight line
Problem:
Given: a straight line h and a point P, not lying on h.
Wanted: a straight line g, running through P and being perpendicular to h.
Solution:
The following window shows the start situation. Maybe, our knowledge about perpendicular bisectors can help us once more? It seems that actual situation is quite similiar to the one on the previous page: if we once more look at the wanted straight line as if it were a perpendicular bisector, then we have to choose the two points A and B on the given line h.
But when P has to be a point of the perpendicular bisector, then the distances of P to A and B must be equal! So A and B must lie on a circle with centre P and any radius. We can get suitable points A and B by intersecting the given line h with a (sufficiently big) circle with centre P.
Now maybe you want to start the normal construction of perpendicular bisector - and you are right, this works too! But in this special situation, there is an easier way to the solution: don't we already know, that P has the same distance from A and from B? Then we can use a circle k1 around A running through P and a cricle k2 around B running through P. P is one of the intersection points of k1 and k2. And the second one (let's name it "Q") also has the same distance from A and B. So P and Q are both (different!) points of the perpendicular bisector g of [AB], and so g can be constructed as the straight line running through P and Q!
It's your turn now to construct the wanted perpendicular line g:
