Geometry Problem

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

Click the figure below to see the problem 750 details.

## Saturday, May 5, 2012

### Problem 750: Complete Quadrilateral, Triangle, Diagonal, Parallel, Similarity, Metric Relations

Labels:
complete quadrilateral,
diagonal,
parallel,
similarity,
triangle

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http://img19.imageshack.us/img19/138/problem750.png

ReplyDeletelet AC cut EF at P ( see sketch)

∆DMA similar to ∆DNF so DA/DF=MA/FN=a/f

∆BMA similar to ∆BGE so BE/BA=GE/AM= e/a

In triangle ∆AEF, applying Ceva’s theorem => PE/PF x DF/DA x BA/BE=1

Replace DA/DF=a/f and BE/BA=e/a we get PE/PF=e/f

In triangle AEF, applying Von Aubel’s theorem => EC/CD=BE/BA+PE/PF

Replace PE/PF=e/f and BE/BA=e/a we get EC/CD=e(f+a)/af

CD/DE=CD/(CD+EC)=af/(af+ef+ea)=c/e

e/c= 1+e/a+e/f

divide both side by e we get 1/c=1/e+1/a+1/f

Draw a line through B parallel to CH to meet EC at K and DA extended at K’.

ReplyDeleteLet BK = b and BK’ = b’

Draw a line through D parallel to CH to meet CF at K and BA extended at L’.

Let DL = d and DL’ = d’

Apply result of Problem 746 to ∆BCD to obtain

c = bd /(b + d),

1/c = 1/b + 1/d ... (i)

Apply result of Problem 746 to ∆BAD to obtain

a = b'd' /(b' + d'),

1/a = 1/b' + 1/d'... (ii)

From similar ∆s EGB and LDB we get

e/d' = BG/BD = GD/BD - 1 = (e/b) - 1,

1/d' = 1/b - 1/e ... (iii)

Exactly in the same manner, we get

1/b' = 1/d - 1/f ... (iv)

Thus by (i) through (iv)

1/a

= (1/d - 1/f) + (1/b - 1/e)

= (1/b + 1/d) - 1/e - 1/f

= 1/c - 1/e - 1/f

Hence 1/c = 1/a + 1/e + 1/f as required

Here is a much simpler Proof:

ReplyDeletec/e = DH/DG, c/f = BH/BN

So BD/c = (BH + DH)/c = BH/c + DH/c = BN/f + DG/e

Next a/e = BM/BG, a/f = MD/DN

So BD/a = (BM + MD)/a = BM/a + DM/a = BG/e + DN/f

Therefore

BD/c - BD/a = (DG - BG)/e + (BN - DN)/f = BD/f + BD/e

Follows 1/c - 1/a = 1/f + 1/e

Hence 1/c = 1/a + 1/e + 1/f