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

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

Click the figure below to enlarge the problem 946.

## Thursday, December 19, 2013

### Geometry Problem 946: Triangle, Quadrilateral, Area, Diagonal, Midpoint, Parallel

Subscribe to:
Post Comments (Atom)

let's take the informal pragmatic approach: if the problem statement does not depend on A, B, C, D position but only on quadrilateral area, it holds for a particular case where, say, B=C=E. In this degenerate configuration, it's easy to see that area(CEFH) = area(ABCD)/4 = 7. Hence this must hold for *any* ABCD ;-)

ReplyDeletebleaug

Let S(ABC) denote the area of ABC, and so on.

ReplyDeleteSince EF//BD//GH, so S(CEHF) = S(CEGF).

Consider quadrilateral CEGF.

CE = 1/2 CB

EG = 1/2 BA

GF = 1/2 DA

FC = 1/2 DC

So CEGF~CBAD.

Hence, S(CEHF) = S(CEGF) = 1/4 S(CBAD) = 7

GH//BD//EF

ReplyDeleteFurther ∆BGF has sides = to half the sides of ∆ABD.

So S(EHF) = S(EGF) = ¼ S(ABD)

Moreover S(ECF) = ¼ S(BCD).

Hence S(CEFH) = S(EHF) + S(ECF) = ¼ {S(ABD) + S(BCD)} = ¼ S(ABCD) =7

Sumith Peiris

Moratuwa

Sri Lanka