## Thursday, September 8, 2016

### Geometry Problem 1256: Parallelogram, Angle Bisector, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1256.

1. https://goo.gl/photos/YYzh4r9rVdU65h2p7

Let CE meet BA extended at N
We have ∠ (BCE)= ∠ (ECD)= ∠ (CED)= ∠ (ANE)= u ( see sketch)
So triangle CDE is isosceles and CD=ED=FG=AB= 28
Triangle GFN is isosceles so FG=FN= 28
Triangle CBN is isosceles so x= BC= BN= 15+28= 43

2. Problem 1256
If CE intersects AB at K then <BCE=<ECD=<CED=<AEK=<AKE .So AB=CD=DE=FG=15+13=28.
And AE=AK=y then triangles KAE and KFG are similar.So AE/FG=KA/KF or y/28=y/y+13 or
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. As [FG]=[ED], [AE]=[GX], being X the intersection of FG and DC.

Since [FBCX] is a parallelogram as well, ^GXC= 180-2(^GCD); So, ^CGX=^GCD and [GX]=[XC]=[BF]=15.

Let P be the intersection of BA and EC. Let [AP]=y.

Since [PAE], [PFG] and [PBC] are all similar, we have:

y/(y+13)=15/(x-15); x= 30 + (13.15/y)

And (y+28)/y= (30+ (13.15/y))/15; y=15

So, x= 30+13 = 43.

DL06

4. Problem 1256
Solution 2
If CE intersect the AB in K then <BCE=<ECD=<DEC=<AEK=<AKE=<FGK so AB=CD=ED=FG=FK=15+13=28 and BC=BK. So AK+AF=FG or AK+13=28 or
AK=15.Therefore x=BC=BK=15+13+15=43.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

5. Let FG meet CD at H.
Tr. CGH is isoceles.
So x - 28 = 15 and hence x = 43

Sumith Peiris
Moratuwa
Sri Lanka

6. We use Barycentric Coordinates.
Let A(1,0,0), B(0,1,0), C(0,0,1)
B + D = A + C (Because ABCD is parallelogram).
Hence, D = A – B + C = (1,-1,1)
F = (15,13,0)/28 (F is on AB, in ratio 13:15)
E is on AB. Let E = A+D(1-q) for some scalar q. Using definition of D, E = (1,-q,q).
Line CE has equation qx + y = 0
[Every line in Barycentric coordinates is in the form ux + vy + wz = 0 for some constant u, v, w. The coefficients are determined by subbing in (x,y,z) and solving for u, v, w.]
G + E = F + D as DEFG is a parallelogram. Since we know E, F and D, we can rearrange this equation to make G = (15/28, q – 15/28, 1 – q)
G is on line CE, so: 15q/28 + q – 15/28 = 0. This means that q = 15/43. G = (15/28, -225/1204, 28/43)
Line GD has equation 559x + 139y – 420z = 0
Make point H which is intersection of GD and BC. This implies that x = 0. H = (0, 420/559, 139/559)
It is obvious that 13H/28 + 15D/28 = G
This means that HG:DG = 15:13
By the angle bisector theorem, it is obvious that HG:DG = HC:CD.
Hence, HC = 420/13
It is also obvious that 420B/559 + 139C/559 = H
Hence, BH:CH = 139:420
This means that HC:CB = 420:559
Finally, CB = (420/13)*(559/420) = 43

Not the cleanest way of doing it, but for fist-brains like me who can't find simple answers, this is epic.

1. TYPO ALERT: When I said "E = A+D(1-q)", I actually meant to say "E = A(1-q)+Dq"