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It's a geometry question. We draw a custom triangle with this features:
1) Angle $\angle A$ is right.
2) the length of AB and AC is more than 9cm.
we cut 5cm from the vector B in AB and name it BE. we cut 9cm from the vector C in AC and name it CD. We connect DE. Then find the midpoint of DE and name it N. We find midpoint of BC and name it M. We connect MN. We want to prove that length of MN is constant.

Image

Is it possible to help me? I'm sorry for Bad English too.
Thanks.

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    What "vector B, vector C" you mean? Perhaps $\;AB, AC\;$ ? – 2017-01-14
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    I mean we cut 5cm in the AB but from B. it means BE is 5cm. – 2017-01-14

3 Answers 3

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You can use coordinate geometry.

Let $E=(0,q), B=(0,q+5), D=(p,0)$ and $C=(p+9,0)$

Then $N=(\frac{p}{2},\frac{q}{2})$ and $M=(\frac{p+9}{2},\frac{q+5}{2})$

$$MN=\sqrt{(\frac{p}{2}-\frac{p+9}{2})^2+(\frac{q}{2}-\frac{q+5}{2})^2}=\frac{\sqrt{106}}{2}$$

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    Hi. It's good way. Is it possible to solve with geometry too? – 2017-01-14
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I mis-read the question and therefore made the following wrong answer. I decided to leave it on because it would be a waste if I have it deleted. The corrected version is appended at the end. Read on.

Let a, b, c be the sides of the given right angled triangle ABC.

enter image description here

$AM = BM = \dfrac {a}{2}$.

Similarly, $AN = \dfrac {\sqrt {(c – 5)^2 + (b – 9)^2}}{2}$

$\angle MAN = \angle MAB - \angle NAE = \tan^{-1} (\dfrac {b}{c}) - \tan^{-1} (\dfrac { b - 9}{c - 5})$

By cosine law, MN is an expression in a, b, and c showing that MN is a constant.

(Corrected version)

Through M, draw line ‘p’ parallel to AC and draw ‘x’ parallel to AB. ‘q’, and ‘y’ are similarly drawn.

enter image description here

Form the parallelogram BENB’ where B’ is a point on ‘y’. Parallelogram NDCC’ is similarly drawn.

Draw ‘r’ and ‘z’ as shown. Let K be the point of their intersection. Then B’NC’K is a rectangle with NK as its diagonal and M happens to be its midpoint. (See the proof shown on the far left.)

Therefore, $NM = \dfrac {NK}{2} = …. = \dfrac {\sqrt {9^2 + 5^2}}{2}$.

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    Hi. I read it again but It's not my question. as I said in my question, It is a custom triangle so the length of MN is not depended on a and b and c in your answer. – 2017-01-14
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    @Amin I mis-read your question and I will delete this post shortly. Will upload the correct answer if I can find one. – 2017-01-15
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    OK. no problem. You can delete it when you want however thanks for naswer. – 2017-01-15
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    @Amin New edition uploaded. – 2017-01-15
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    Thanks for answer. I check it. It's very good. You have awesome imagination. You can remove your first answer to make your answer clearly. Good luck. – 2017-01-15
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Thanks everyone for answers. I find other way to solve this question and I want to tell that here.
We have this information: Image

In the $\triangle DEM$ we use Stewart's theorem: $MN^2 = \frac{2ME^2 + 2MD^2 - 4t^2}{4}$
Then with law of cosines in the $\triangle MEC$ and $\triangle BDM$ and use them in the Stewart's theorem:

$\frac{1}{2}(81 + a^2 - 18a\cos \alpha) + \frac{1}{2}(25+ a^2 - 10a\cos (90 - \alpha)) - \frac{4t^2}{4}$ =
$53 + a^2 - 9a\cos \alpha - 5a \sin \alpha - \frac{4t^2}{4}$ =
$53 + a^2 - 9a(\frac{b}{2a}) - 5a(\frac{c}{2a}) - \frac{(b-9)^2 + (c - 5)^2}{4}$ =
$53 + a^2 - \frac{9b}{2} - \frac{5c}{2} - \frac{b^2 + c^2 + 81 + 25 - 18b - 10c}{4}$
We know that $b^2 + c^2 = 4a^2$ so:

$53 + a^2 - \frac{9b}{2} - \frac{5c}{2} - a^2 - \frac{53}{2} + \frac{9b}{2} + \frac{5c}{2} = \frac{106}{4}$
so $MN = \frac{\sqrt{106}}{2}$

However thanks for everyone.