## Friday, August 29, 2014

### Geometry Problem 1040: Isosceles Triangle, Transversal Line, Metric Relations

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 1040. 1. Using mass point geometry. Let m(P)=mass of point P.

Let AB=BC=a.

If m(B)=15, then m(B)/5=m(A)/(a-5), so m(A)=3a-15.
Also, we have m(B)/3=m(C)/(a-3), so m(C)=5a-15.

Hence, m(D)=m(A)+m(B)=3a; m(E)=m(B)+m(C)=5a.
Since m(E)=m(D)+m(F), thus m(F)=2a.

Now m(D):m(F)=3:2, but also m(D):m(F)=8:x.
So 8/x=3/2, and finally x=16/3.

2. for the theorem of parallels we have 5:3 = (X+8):8 and therefore X= 16/3

3. 8/(x+8)=[ECF]/[DCF]=[EAF]/[DAF]=([EAF]-[ECF])/([DAF]-[DCF])=[EAC]/[DCA]=3/5
40=3x+24, x=16/3

4. Use menelaus theorem on triangle ADF and line BEC and then on triangle ABC and line DEF

5. EG//CA drawn,with G on AB.
`.`BG=BE.
.'.GA=BA-BG=BC-BE=EC=3 & so DG=2
InΔDAF,GE//AF .'.2/3=x/8 =>x=16/3

6. by menelaus theorem, triangle BDE and line ACF
(DA/AB)(BC/CE)(EF/FD) = 1
(5/AB)(BC/3)(8/(8+x)) = 1
but AB = BC
40=3(8+x) --> x = 16/3

7. 2 simple ways of doing it

Draw a parellel to AC thro D then 2/3 = x/8

Or draw altitudes h1 and h2 from D and E and from similar Tr.s agIn (x+8)/8 = h1/h2 = 5/3

From both approaches x = 16/3

Sumith Peiris
Moratuwa
Sri Lanka

8. 2 more simple ways

Get similar Tr.s by either drawing a parellel to BC thro D or a parellel to AB thro E