# Thread: Construct vertex D of an acute angled triangle

1. You know the formula for the area of a triangle. Why don't you describe your ideas on the location of $D$?

Both the triangles should be on the same base and between same pair of parallel lines

I am not sure whether it correct but here are my ideas

I think It has got to do something with parallelograms, I guess?

Many Thanks

3. Originally Posted by mathlearn
Both the triangles should be on the same base and between same pair of parallel lines
Yes, but the problem statement also stipulates that $CA=CD$. So you should draw the line through $C$ that is parallel to $AB$ and then mark $D$ on that line so that $CA=CD$. To draw a parallel line through $C$, see .

4. It has to do with the fact that the area of a triangle is "(1/2) base times height". Since the two triangles will have the same base, their heights must also be the same. That is the reason for drawing the line, through C, parallel to AB. The further condition is that "CA= CD". Set one leg of a pair of compasses at C, set the other on A, and draw an arc with center at C and radius CA. D is where the line and arc intersect.