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I've tried to use the distances between the points (not the center of the squares), and I keep getting stuck. Help!

 Mar 15, 2020
 #1
avatar+109731 
0

Visually I can see it is true but I'm stuck too.

 

Here is my GeoGebra construction. Hopefully you will be able to access it properly.

Although I suppose it will not be much help anyway.

(It is interactive)

 

https://www.geogebra.org/classic/yzxphpyq

 Mar 15, 2020
edited by Melody  Mar 15, 2020
 #2
avatar+109731 
0

I'd be interested to see the solution for this one too.   

 Mar 15, 2020
 #3
avatar+25265 
+3

Please help!

 

a)

Find expressions for p,q,r and s in terms of a,b,c and d

\(\begin{array}{|rcll|} \hline p &=& \dfrac{a+b}{2} + \left(\dfrac{b-a}{2}\right)_\perp \\ 2p &=& (a+b) + (b-a)_\perp \qquad \text{rotate clockwise $\times (-i)$ }\\ 2p &=& (a+b) + (b-a)\times (-i) \\ \mathbf{2p} &=& \mathbf{(a+b) + (a-b)i}\\ \text{similarly} \quad \\ \mathbf{2q} &=& \mathbf{(b+c) + (b-c)i}\\ \mathbf{2r} &=& \mathbf{(c+d) + (c-d)i}\\ \mathbf{2s} &=& \mathbf{(d+a) + (d-a)i}\\ \hline \end{array} \)

 

b)
Prove that the line segment between p and r
is perpendicular  to the line segment between q and s

\(\begin{array}{|rcll|} \hline 2rp &=& 2(r-p) \\ &=& 2r-2p \\ &=& (c+d)+(c-d)i-(a+b)-(a-b)i \\ &=& (c+d)-(a+b)+(c-d)i-(a-b)i \qquad (1) \\\\ 2sq &=& 2(s-q) \\ &=& 2s-2q \\ &=& (d+a)+(d-a)i-(b+c)-(b-c)i \\ &=& (d+a)-(b+c)+(d-a)i-(b-c)i \qquad (2) \\\\ \hline \mathbf{2rp_\perp} &=& \mathbf{2sq\ ?} \\ 2rp &=& (c+d)-(a+b)+(c-d)i-(a-b)i \qquad \text{rotate counter clockwise $\times i$ } \\ 2rp_\perp &=& \Big((c+d)-(a+b)+(c-d)i-(a-b)i\Big) \times i \\ 2rp_\perp &=& (c+d)i-(a+b)i+(c-d)i^2-(a-b)i^2\qquad i^2=-1 \\ 2rp_\perp &=& (c+d)i-(a+b)i-(c-d)+(a-b) \\ 2rp_\perp &=& (d+a)-(b+c)+(d-a)i-(b-c)i \\ 2rp_\perp &=& 2sq \checkmark \\ \hline \end{array} \)

 

Prove that the line segment between p and r
equal in length to the line segment between q and s

\(\small{ \begin{array}{|rcll|} \hline && \Big[(c+d)-(a+b)\Big]^2 +\Big[(c-d)-(a-b)\Big]^2 \\ &=&\Big[(d+a)-(b+c)\Big]^2 +\Big[(d-a)-(b-c)\Big]^2 \\\\ && (c+d)^2+(a+b)^2-2(c+d)(a+b) \\ && +(c-d)^2-(a-b)^2-2(c-d)(a-b) \\ &=& (d+a)^2+(b+c)^2-2(d+a)(b+c) \\ && +(d-a)^2+(b-c)^2-2(d-a)(b-c) \\\\ &&c^2+d^2+2cd+a^2+b^2+2ab \\ && -2ac-2bc-2da-2bd \\ && +c^2+d^2-2cd+a^2+b^2 \\ && -2ab-2ac+2bc+2da-2db \\ &=& d^2+a^2+2da+b^2+c^2+2bc \\ && -2db-2dc-2ab-2ac \\ && +d^2+a^2-2da + b^2+c^2-2bc \\ && -2db+2dc+2ab-2ac \\\\ &&c^2+d^2+a^2+b^2-2ac-2db \\ && +c^2+d^2+a^2+b^2-2ac-2db \\ &=& d^2+a^2+b^2+c^2 -2db-2ac \\ && +d^2+a^2+ b^2+c^2-2db-2ac \\\\ && 2c^2+2d^2+2a^2+2b^2-4ac-4db \\ &=& 2c^2+2d^2+2a^2+2b^2-4ac-4db\ \checkmark \\ \hline \end{array} }\)

 

laugh

 Mar 15, 2020
 #4
avatar+117 
+2

This is known as Van Aubel's Theorem.

 Mar 15, 2020
 #5
avatar+109731 
+1

Thanks Heureka and Tiggsy.

Now I have some homework   laugh

 Mar 16, 2020

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