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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.

enter image description here

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.

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    $\begingroup$ imgur.com/a/czyAJEl $\endgroup$
    – user967265
    Sep 22, 2021 at 6:20
  • $\begingroup$ Pardon for the innaccuracy of the diagram. $\endgroup$
    – user967265
    Sep 22, 2021 at 6:20
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    $\begingroup$ 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. $\endgroup$ Sep 22, 2021 at 7:12
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    $\begingroup$ Gah, I didn't see a mistake until after the edit window. $K$ is at $(x_1, \sqrt 3)$, not $1/\sqrt 3$. $\endgroup$ Sep 22, 2021 at 7:26
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    $\begingroup$ HINT. Point $C$ must be on the left of $D$. It turns out that $C$ is the midpoint of $BD$. $\endgroup$ Sep 22, 2021 at 10:37

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enter image description here

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?

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  • $\begingroup$ 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. $\endgroup$ Sep 22, 2021 at 20:06
  • $\begingroup$ @EricSnyder Thanks a lot. You correctly noted that the triangle is obtuse. As for the image, I made it with AutoCAD. $\endgroup$
    – Saeed
    Sep 23, 2021 at 0:29

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