Geometry Problem 1524. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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Let AE,DF extended meet at X and let BE,CF extended meet at Y.

ReplyDeleteLet BE = DF = p and AE = FC = q. Let FY = u and EY = v

< ABC & < BAD are supplementary and so < ABE & < BAE are complementary

Hence EX FY is a Rectangle

From similar triangles ABE & BYC,

(p+v)/p = (q+u)/q = 10/6

So u = 2q/3 and v = 2p/3

Applying Pythagoras to Triangles ABE & EYF,

EF^2 = u^2 + v^2 = (2/3)^2. (p^2 + q^2) = (2/3)^2. (6^2) = (4/9)*36 = 16

Therefore EF = 4

Sumith Peiris

Moratuwa

Sri Lanka

Extend DF to P => BP=10 - 6 = EF

ReplyDeleteHow is BP = EF?

DeleteOn what basis is BP = EF?

Delete( BE = PF, & BE // PF ) => BPFE parallelogram

DeleteDear Peter - could you explain how you concluded that points M,E,F,N are collinear which is essential for your proof to be valid?

ReplyDeleteRegards

Sumith

Since M and N are midpoints of AB and CD => MN//BC

ReplyDeleteME//BC and NF //BC => M,E, F ,N are collinear

Thanks

Delete