Online Geometry theorems, problems, solutions, and related topics.
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.
https://goo.gl/photos/YYzh4r9rVdU65h2p7Let CE meet BA extended at NWe have ∠ (BCE)= ∠ (ECD)= ∠ (CED)= ∠ (ANE)= u ( see sketch)So triangle CDE is isosceles and CD=ED=FG=AB= 28Triangle GFN is isosceles so FG=FN= 28Triangle CBN is isosceles so x= BC= BN= 15+28= 43
Problem 1256If 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 ory+13=28 so y=15.But x=BC=AD=y+28=15+28=43.APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
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=15So, x= 30+13 = 43.DL06
Problem 1256Solution 2If 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 orAK=15.Therefore x=BC=BK=15+13+15=43.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Let FG meet CD at H. Tr. CGH is isoceles.So x - 28 = 15 and hence x = 43Sumith PeirisMoratuwaSri Lanka
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 = 0Make 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 = GThis means that HG:DG = 15:13By the angle bisector theorem, it is obvious that HG:DG = HC:CD.Hence, HC = 420/13It is also obvious that 420B/559 + 139C/559 = HHence, BH:CH = 139:420This means that HC:CB = 420:559Finally, CB = (420/13)*(559/420) = 43Not the cleanest way of doing it, but for fist-brains like me who can't find simple answers, this is epic.
TYPO ALERT: When I said "E = A+D(1-q)", I actually meant to say "E = A(1-q)+Dq"