## Tuesday, January 3, 2017

### Geometry Problem 1302: Triangle, Median, Circles, Midpoint, Congruence

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

Details: Click on the figure below. 1. connect AE, connect CF, AE//CF;
extend 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

2. https://goo.gl/photos/9ENrkK4G1b2hs1BCA

Let 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

3. Problem 1302
Is <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

4. Solution 1

Let 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

5. Solution 2

Draw 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

6. Solution 3

Let FM extended meet AE at L

∆MAL ≡ ∆MFC (ASA)

So LM = MF = ME since < LEF = 90

Sumith Peiris
Moratuwa
Sri Lanka

7. 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.

Best regards

8. AE and CF are perpendicular to BD, Since M is midpoint of AC it lies on perpendicular bisector of EF. Hence ME=MF