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

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

Details: Click on the figure below.

## 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