# Constructing triangle $\triangle ABC$ given median $AM$ and angles $\angle BAM, \angle CAM$

Constructing triangle $\triangle ABC$ given median $AM$ and angles $\angle BAM, \angle CAM$

I start with the median $AM$. Since $\angle BAM, \angle CAM$ are known I can construct them. So I have point $A$, line segment $AM$ and $2$ rays starting from A where $\angle BAM, \angle CAM$ are those from the hypothesis. If only I could find a way to construct a line where $BM = MC$ I would be done. I can't figure out how to do that.

• did you tried draw a circle with center in $M$ ? – Luis Felipe Nov 5 '15 at 14:31
• I mean, the diameter must to be your side $BC$. $B,C$ must be the intersection of the circunference with the rays, think on it. – Luis Felipe Nov 5 '15 at 14:32
• Yes, how does it help me, though? Suppose I have a circle with center $M$ and radius $|AM|$ it's not onto $B$ or $C$ since $\angle BAC$ is not a right angle. – kg_b Nov 5 '15 at 14:37
• ok, i'll answer your question. let me draw. please wait – Luis Felipe Nov 5 '15 at 14:40
• I trough use some Euler Line to solve this but i'm mistaken, here you have the fast answer. – Luis Felipe Nov 5 '15 at 15:06

## 2 Answers The black lines are your dates. at first we draw green lines which are parallels to the rays in the point $M$, so we get the points $J,K$. with center in $J$ and radio $AJ$ we draw the red circle. similar with the point $K$ so we get the purple points $B,C$ and the segment $BC$ is your goal.

• What program did you use for drawing? – kg_b Nov 5 '15 at 15:46
• its name is geogebra – Luis Felipe Nov 5 '15 at 15:46

The symmetric of $B$ with respect to $M$ is $C$, hence in order to find our triangle it is enough to intersect the $AC$ line with the symmetric of the $AB$ line with respect to $M$. 