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In a trapezoid ABCD is M = mi $[DB]$ and N= mi $[AC]$

Proof that $|MN| = \frac {1}{2} (|DC| - |AB|)$

I've already proven this analytically with points A(2,4); B(6,4); C(10,0); D(0,0) and in the general sense just by using slopes. But I don't know how to proof this geometrically.

Can someone give me a hint please?

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  • $\begingroup$ Are you sure your first statement is correct? Do you not mean $\text{midpoint}(AB)$ and $\text{midpoint}(CD)$? $AC$ is a diagonal non-edge, from my understanding. $\endgroup$ – Axoren Aug 31 '16 at 13:26
  • $\begingroup$ Yes, that is what I meant with M= mi $[AB]$ (= the midpoint of $[AB]$) $\endgroup$ – Anonymous196 Aug 31 '16 at 13:28
  • $\begingroup$ @AnonymousI $MN$ is not a parallel segment as defined. It is a diagonal. Something about your original statment is lacking or incorrect. $\endgroup$ – Axoren Aug 31 '16 at 13:32
  • $\begingroup$ No, it does because if you calculate the midpoints M and N you will get resp. co(M)= (3,2) and co(N)= (6,2). And then you calculate the distance between M and N and then you get 3 which is exactly what the formula says since the distance of |DC|= 10 and |AB|= 4 $\endgroup$ – Anonymous196 Aug 31 '16 at 13:33
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    $\begingroup$ Now I see I've made some errors above. Now it should be correct. $\endgroup$ – Anonymous196 Aug 31 '16 at 13:39
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Let $P$ and $Q$ be the intersections of $MN$ with $AD$ and $BC$.
By Thales' theorem $MN\parallel AB\parallel CD$ and $$ PM=NQ=\frac{1}{2}CD,\qquad PQ=\frac{AB+CD}{2} $$ hence it follows that: $$ MN = PQ-PM-NQ = \frac{AB+CD}{2}-CD = \color{red}{\frac{AB-CD}{2}} $$ as wanted.

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  • $\begingroup$ I'm sorry sir, I've made some errors above in the information section. $\endgroup$ – Anonymous196 Aug 31 '16 at 13:41
  • $\begingroup$ Hate to be that guy but you've got $M$ and $N$ the wrong way round (not that it makes a difference). $\endgroup$ – Jam Sep 5 '16 at 18:13
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    $\begingroup$ Yes, you're right, I switched M and N, but that is irrelevant here. $\endgroup$ – Jack D'Aurizio Sep 5 '16 at 18:29

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