## Saturday, May 5, 2012

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

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

Click the figure below to see the problem 750 details.

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

let 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

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

3. Here is a much simpler Proof:
c/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

4. Apply Menelaus on ABF
=> (a+e/e)*(BC/CF)*(f/a)=1
=> BC/CF=a.e/f(a+e)
=> BC/BF=a.e/(a.e+f(a+e))=c/f
=>1/c=a.e+f(a+e)/a.e.f
Hence 1/c=1/a+1/e+1/f