Geometry Problem

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

Click the figure below to see the complete problem 919.

## Saturday, September 7, 2013

### Geometry Problem 919: Rhombus, Triangle, Perpendicular, 90 Degrees, Congruence

Labels:
90,
congruence,
perpendicular,
rhombus,
triangle

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Let a denote the length of each side of the rhombus.

ReplyDeleteBD bisects angle ABC

Let h be the distance of D from BA or BC.

Clearly MG = h = HP.

(i.e.)FG - FM = EP - EH

So FG + EH = EP + FM

Rewrite the statement to FG-FM = EP-EN

ReplyDeleteBoth sides equals to the height of the rhombus, thus proved.

FM + EN = FM + EN, GM = PN from being altitude of rombus

ReplyDelete(GM + MF) + EN = FM + (PN + NE)

GF + EN = EP + FM

To Antonio: I think the problem I suggested a few days ago is more interesting that these problems. However, it is your right to make the selection.

Erina-NJ

Hi Erina, your problem has been published at

Deletegogeometry.com/school-college/p920-triangle-angle-10-20-30-degrees.htm

Thanks