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