Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, August 11, 2018

### Geometry Problem 1378: Parallelogram, Diagonal perpendicular to a side, Distances from a Point to Vertices, Measurement

Labels:
diagonal,
distance,
geometry problem,
parallelogram,
perpendicular

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ReplyDeletelet AC meet BD at M

FM is the median of triangles AFC and BFD

We have AF^2+FC^2= a^2+c^2= 2( MA^2+MF^2)

And FB^2+FD^2=b^2+d^2= 2(MB^2+MF^2)

In right triangle ABM we have MA^2-MB^2= AB^2= e^2

So a^2+c^2= b^2+d^2+2.e^2

Apply Apollonius to Tr.s AFC & BFD remembering that BD & AC bisect each other

ReplyDeleteThen apply Pythagoras to ABX, X being where the diagonals intersect

Let M be the midpoint of BD.

ReplyDeleteJoin MF and apply apollonius to the triangle BFD, then we get :

b^2+d^2=2MF^2+BD^2/2

which we rewrite to:

(1).BD^2=2b^2+2d^2-4MF^2

If we apply apollonius to the triangle AFC we can get the following equation :

(2).AC^2=2a^2+2c^2-4MF^2

Applying pythagoras to ABD we get :

AD^2=AB^2+BD^2

using equations 1 and 2 we get:

(3).AD^2=e^2+2b^2+2d^2-4MF^2

If we apply the parallelogram law :

AC^2+BD^2=2AB^2+2AD^2

using equations 1,2 and 3 :

2a^2+2c^2+2b^2+2d^2-8MF^2=4e^2+4b^2+4d^2-8MF^2

2a^2+2c^2=4e^2++2b^2+2d^2

a^2+c^2=2e^2+b^2+d^2