How Do You Construct a Triangle with Given Side, Angle, and Median Length?

[wubie]

Hello,

I am having trouble understanding the solution to a problem. It should be straightforward but as usual it is not straightforward to me.

Question:

Construct a triangle ABC given side a, angle B, and the length of the median from C.

Now, I first construct the side a (side BC). I then construct angle B (say XBC). Now as far as I understand it the median is a segment that joins the vertex to the midpoint of the opposite side.

In the solution given to me, it says to let m be the length of the median. With the center C and radius m, draw a circle cutting the ray BX at M (Depending upon the length m, there may be 1 or 2 possible points M). This step I do not get.

If length m is the length of the median, using C as the center, how would I possibly get 2 points? I can only visualize one point. And that one point is tangent to the ray BX. And that tangent point is not very precise.

Could someone spell it out for me? I must be missing something elementry.

Any help is appreciated.

Thankyou.

 

[HallsofIvy]

Start as you said: Draw side a of the given length. At one end is angle B, at the other, angle C. Construct the given angle B and extend the ray from angle B (this is side c- we don't its length).

Now the complicated part: We know the length of the median through C, we just don't know its direction! Construct the circle with center at C, radius the given length of the median. The points on this circle represent all possible endpoints for the median. Of course, the other end of the median has to be on line c- that is it must be where the circle and the line intersect. Actually, there will, in general, be two solutions. The circle constructed may cut the line in two different points.

Select either of the two points where the circle cuts the line (if there are two). That point is the midpoint of side c. Now just use a compass to duplicate the length from B to that midpoint on c on the other side of the midpoint. The result is point A of the triangle.

 

[wubie]

Thanks for you help Ivy. I know where I went wrong now.

I unconsciously assumed that the triangle I was constructing had the same properties as the properties of an equilateral triangle. But at the same time I was thinking that the triangle could be any triangle. Of course I had a conflict of properties there. But at the time I couldn't see why.

For some reason I was thinking that the median had to be perpendicular to the side it was intersecting in all cases of triangles. I know that this is NOT the case unless the median is also the bisector of the angle opposite to the side the median intersects. And I can only think of this being true for two triangles - an equilateral triangle, and the unequal angle of an isosceles triangle. I don't know why I kept on insisting to myself that the median had to be perpendicular to the side it intersected even though I was not constructing an isosceles or equilateral. Just one of those things that happens when one is tired I guess.

Once again, thanks for your help.

Sincerely