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I would like someone to verify that I am solving this problem correctly. I do not remember the theorem that allows me to make the two halves of the triangle proportional. Because (h1/h2 = h1/h2) Triangles are proportional?

Here is the problem: enter image description here

My work:

enter image description here

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  • $\begingroup$ It is not clear that what you have done in your soulution. $\endgroup$
    – Ömer
    Nov 20, 2013 at 21:21
  • $\begingroup$ @Ömer I appreciate your answer, but I would like to know whether my logic/method is correct. $\endgroup$ Nov 21, 2013 at 6:12

1 Answer 1

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$$\frac{Area(BML)}{Area(BCM)}=\frac{LM}{MC} \Rightarrow \frac{5}{10}=\frac{LM}{MC}$$ $$\frac{Area(MCK)}{Area(BCM)}=\frac{KM}{MB} \Rightarrow \frac{8}{10}=\frac{KM}{MB}$$ enter image description here

Let's say the area of $AMK$ be $2A$ then from $\frac{Area(ALM)}{Area(AMC)}=\frac{LM}{MC}$, the area of $ALM$ will be $4+A$. Now

$$\frac{Area(AMK)}{Area(ABM)}=\frac{KM}{MB} \Rightarrow \frac{2A}{5+4+A}=\frac{4a}{5a}\\\Rightarrow 10A=36+4A\Rightarrow A=6 \Rightarrow S=4+A+2A=22$$

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  • $\begingroup$ Thanks for your help! I got the same answer with my solution as well, I just wanted to make sure it wasn't a coincidence. $\endgroup$ Nov 20, 2013 at 22:18
  • $\begingroup$ You are welcome. $\endgroup$
    – Ömer
    Nov 20, 2013 at 22:19

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