# Finding the length of a triangle given one side and the ratio the median and angle bisector cut the altitude.

A median BK, an angular bisector BE and an altitude AD are drawn in a triangle ABC. Find the side AC if it is known that the line BK and BE divide the line segment AD into 3 equal parts and if AB = 4cm. Find the length of AC.

Currently, I am able to work out that BD=2 by angle bisector theorem and that,since,sin BAD= 1/2; BAD=30 and angle ABC=60 but after that I'm unable to work it out further.

• imgur.com/a/czyAJEl
– user967265
Commented Sep 22, 2021 at 6:20
• Pardon for the innaccuracy of the diagram.
– user967265
Commented Sep 22, 2021 at 6:20
• I think converting to a coordinate system works. Setting $B$ as the origin, $A$ is at $(2,2 \sqrt 3)$, and $M$ is at $(2,4/ \sqrt 3)$. Then $AK$ has slope $2/ \sqrt 3$. $K$ must be at $(x_1, 1/ \sqrt 3)$, since its y-coordinate must be halfway between the y-coordinate of $A$ and $C$, which is at $(2x_1,0)$. Now find the point on $AK$ that has that y-coordinate. I suspect, though I haven't gone all the way through, that the triangle is obtuse. Commented Sep 22, 2021 at 7:12
• Gah, I didn't see a mistake until after the edit window. $K$ is at $(x_1, \sqrt 3)$, not $1/\sqrt 3$. Commented Sep 22, 2021 at 7:26
• HINT. Point $C$ must be on the left of $D$. It turns out that $C$ is the midpoint of $BD$. Commented Sep 22, 2021 at 10:37

I can help you with the construction, and you should be able to do the rest. Forget about the length of $$AB$$ for now.

1. Draw altitude $$AH$$ , draw the baseline (where points $$B$$ and $$C$$ should lie), and draw the midline (where point $$K$$ should lie). Mark points $$X$$ and $$Y$$, which divide $$AH$$ into 3 equal segments.

2. Draw circle $$c$$ with center $$X$$ and radius $$XH$$.

3. $$X$$ is on the bisector of $$\widehat{ABC}$$, so $$AB$$ should be tangent to circle $$c$$. Draw $$AB$$.

4. Draw $$BY$$. It meets the midline at $$K$$, the midpoint of $$AC$$.

5. Draw $$AK$$ and extend it to meet the baseline at $$C$$.

Now, can you do the rest?

• Oh, this is beautiful. It's where my algebra comment leads, but without the algebra. I love seeing a good construction, and hopefully OP can follow it readily. Also a tip of the hat to Geogebra, I assume. Commented Sep 22, 2021 at 20:06
• @EricSnyder Thanks a lot. You correctly noted that the triangle is obtuse. As for the image, I made it with AutoCAD. Commented Sep 23, 2021 at 0:29