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# hard bisector geometry help

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In triangle $$ABC$$$$AB=5$$$$BC=8$$, and the length of median $$AM$$ is 4. Find $$AC$$.

Thank you :D

Jan 31, 2022

#1
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I got $$\sqrt{39}$$ with long equations and polynomials, can someone double check thanks!

Jan 31, 2022
#2
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Here is another method, using the cosine law twice only.

Consider triangle ABM, AB=5, AM=4 and BM = 4 (Because the median bisects the opposite side to the vertex).

So applying cosine rule gives you the angle ABC (which is the same as ABM)

Then, consider the triangle ABC, we have, AB=5 and BC=8 and we have the angle ABC.

Then apply cosine rule, yielding AC to be $$\sqrt{39}$$

Guest Jan 31, 2022
#3
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Thanks for double checking, also I think you can use AC as the diameter of a circle, then knowing ABC is a right triangle and then using the pythagorean theorem, thanks anyway :D

proyaop  Jan 31, 2022
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Wow! This is really a neat solution!

Guest Jan 31, 2022