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