## Saturday, August 11, 2018

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. https://photos.app.goo.gl/oB6yLqwymZStn8EP8
let 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

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

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

3. Let M be the midpoint of BD.
Join 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 :

using equations 1 and 2 we get: