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

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

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## Thursday, July 9, 2015

### Geometry Problem 1131: Triangle Area, Two Cevians, Equal Product of Areas

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Let BF cut ED and AC at H and G

ReplyDeleteObserve that H and G are harmonic conjugate points of F and B

So GF/GB=HF/HB

But S3/S= GF/GB and S2/S3= HF/HB

The result will follow.

Lets compute the ratios S/S3 and S1/S2 and prove they are equal:

ReplyDeleteS3/ACD = AF/AD (triangles with same height)

ACD/S = CD/BC => ACD = (CD/BC)*S

S3 = (AF/AD) * ACD = (AF/AD) * (CD/BC) * S

S / S3 = (AD / AF) * (BC / CD)

S2 / ADE = DF / AD

ADE / S1 = AE / BE => ADE = (AE / BE) * S1

S2 = (DF/AD) * ADE = (FD / AD) * (AE / BE) * S1

S1 / S2 = (AD / DF) * (BE / AE)

Next use Menelaus' theorem for the triangle ABD and the line CFE:

( BE / AE) * ( AF / DF ) * ( CD / BC ) = 1

( BE / AE) * ( AF / DF ) = BC / CD (multiply both sides by AD/AF)

( BE / AE) * ( AD / DF ) = (BC / CD) * (AD / AF)

These are exactly the expressions above for S1 / S2 and S / S3

Therefore S1 / S2 = S / S3

Hence S.S2 = S1.S3

From tr EDB and tr EDC => S1/(S2+S5) = BD/DC

ReplyDeletefrom tr ADB and tr ADC => (S1+S2+S4)/(S3+S5) = BD/DC

or S1/(S2+S5) = (S1+S2+S4)/(S3+S5)

but S4 S5 = S2 S3 => the result