See complete Problem 90

Quadrilateral and Triangle areas, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Geometry Problem 90

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## Monday, May 19, 2008

###
Geometry Problem 90

## Link List

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 90

Quadrilateral and Triangle areas, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

Let X be the mid-pt of BC,

ReplyDeleteXM//BA and XN//CD,

[EBM]+[ENC]=[EBX]+[EXC]=[EBC]

S_1=[BMNC]=1/2[BANC]=1/2*S/2=S/4

Area triangle ABC is (ABC) then (BMD)=(ABD)-(ABM)-(AMD).But (MEN)=(MED)-(NED)-(NMD)=(MEC)+(MCD)-(BED)/2-(BMD)/2=(AEC)/2+(ACD)/2-(BED)/2-(ABD)/2+(ABM)/2+(AMD)/2=(AED)/2-(BED)/2-

ReplyDelete(ABD)/2+(ABM)/2+(AMD)/2=(ABD)/2-(ABD)/2+(ABM)/2+(AMD)/2=(ABC)/4+(ACD)/4=

(ABCD)/4.