Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Tuesday, January 3, 2017
Geometry Problem 1302: Triangle, Median, Circles, Midpoint, Congruence
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connect AE, connect CF, AE//CF;
ReplyDeleteextend EM, extend CF, cross at G,
<CAE=<ACF =<ACG, <AME=<GMC, AM=MC
triangle AEM = triangleGMC,
so GM=ME, triangle EGF = right angle, so EM=MF
https://goo.gl/photos/9ENrkK4G1b2hs1BCA
ReplyDeleteLet G and H are the midpoints of AD and DC
We have AM=GH= ½.AC
GM= AM-GA=GH-GD= DH
So GE=GD=MH
And GM=DH=FH
Triangles EGD and DHF are isosceles => ∠EGD=∠DHF
Triangles GME congruent to HFM ( case SAS)=> ME=MF
Problem 1302
ReplyDeleteIs <AED=90=<DFC then AE//CF or AEEF is trapezoid.Let N midpoint in EF, so MN//AE//CF then MN is perpendicular bisector EF.Therefore ME=MF.
APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE
Solution 1
ReplyDeleteLet the circles be P (larger) and Q (smaller) with radii p and q respectively.
So AM = p+q and hence PM = q and QM = p.
Therefore easily ∆MPE ≡ ∆FQM (SAS) and hence,
ME = MF
Sumith Peiris
Moratuwa
Sri Lanka
Solution 2
ReplyDeleteDraw MN // AE//FC, N on FE
Let EN = u, ND = v and DF = w
From similar ∆s,
u/v = (p+q)/(p-q) and w/v = 2q/(p-q)
Subtracting, u/v – w/v = (p+q-2q)/(p-q) = 1
Hence u-w =v and so EN = NF implying that ME = MF
Sumith Peiris
Moratuwa
Sri Lanka
Solution 3
ReplyDeleteLet FM extended meet AE at L
∆MAL ≡ ∆MFC (ASA)
So LM = MF = ME since < LEF = 90
Sumith Peiris
Moratuwa
Sri Lanka
See that EF is the projection of AC onto BD, wherefrom the solution: the midpoint of the segment is equally apart of the ends of the projection.
ReplyDeleteBest regards
AE and CF are perpendicular to BD, Since M is midpoint of AC it lies on perpendicular bisector of EF. Hence ME=MF
ReplyDelete