Question P.414: Another solution to Exercise 329: to draw through a given po......
Another solution to Exercise 329: to draw through a given point inside an angle a secant which forms, with the sides of the angle, a triangle with given area. Construct first the parallelogram with a vertex at the given point, and two sides on the sides of the angle. This parallelogram cuts from the required triangle two partial triangles whose sum is known. Reduce then to the question in Exercise 216.
Suppose (fig. t414) we must draw secant MON through point O inside angle {\widehat{X S Y}}, such that (using absolute value for area) |MNS| is equal to some value ∆. We first construct parallelogram AOBS. Then, if we drop perpendiculars OP, OQ from O to SY, SX respectively, we have AM · OP + BN · OQ = 2(∆ − |AOBS|.
We now construct segment {{B}}B^{\prime} such that |B O B^{\prime}| = ∆ − |AOBS|. Then we have AM · OP + BN · OQ = {{B}}B^{\prime}· OQ, or (subtracting 2|BON| from both sides) AM · OP = NB^{\prime}· OQ.
So we are led to the situation where we are given points A,\;\;B^{\prime}, and we must construct a secant MON such that A M:B^{\prime}N=O Q:O P. This is the situation of exercise 216.
Note. Actually, we do not need AOBS to be a parallelogram. It can be any quadrilateral such that ∆ −2|AOBS| is non-negative, and the argument still holds.
