The value of segment GH in the square below

geometry

For reference: If ABCD and EFGH are squares, HI = 2 and GN=$\sqrt5$ , calculate EH.

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My progress:

If point I were given as the midpoint of $AD$ the exercise would be quite easy $AE=4=AI, EH = \sqrt{(4^2+2^2)} = \sqrt20 = 2\sqrt5=GH$
so I think the idea is to demonstrate that it is the midpoint. I drew some auxiliary lines…

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Best Answer

Let the vertex of right angled triangle it's hypotenuse is $GH=\sqrt 5=\sqrt{2^2+1^2}$ be M, then:

$\triangle BEF\approx \triangle MGH$

and we have:

$\frac{GN=\sqrt 5}{BF}=\frac {MN=\frac 12}{BE=1}$

which gives :

$EH=EF=2\sqrt 5$

This is true if $BF=\frac{BE}2=HI$

The ratio of sides adjacent to the rught angle is $\frac 12$ in all similar triangles MGN, BFE and HID.

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