triangle ABC. Select a point P inside the triangle and draw lines
AP, BP, and CP extended to their intersections with the opposite
sides in points D, E, and F respectively.
the point P around and explore the lengths of AF, BD, CE, EA,
FB, and DC.
the point P and calculate the same relationship:
the point P one more time and again calculate the relationship:
When P is inside the triangle the relationship (AF*BD*CE)/(EA*FB*DC)=1.
Now we will
prove the conjecture we just made.
we just made is called Ceva's Thoerem and
it states that:
points F, D and E are on the sides AB, BC and AC of a triangle
then the lines AD, BE and CF are concurrent if and only if the
product of the ratios
the lines BE and CF beyond the triangle and draw a line through
A and parallel to BC. Mark the points where the the extended lines
cross the parallel line.
are several pairs of similiar triangles which give us the following
EBC and EAY are similiar triangles
we get the ratio: (EC/EA)=(BC/YA)
FBC and FAX are similiar triangles
we get the ratio: (FA/FB)=(AX/BC)
XAP and CDP are similiar triangles
we get the ratio: (XA/CD)=(PA/PD)
BDP and YAP are similiar triangles
we get the ratio: (BD/YA)=(PD/PA)
we mulitply the ratios together, we get:
when we simplify we get:
mulitple both sides by (GA/AX)
our final formula is:
have proved our conjecture.
would happen if P is outside the triangle? You can explore by
using the GSP
Show that when P is inside the triangle ABC, the ratio of the
areas of the triangle ABC and triangle DEF is always greater than
or equal to 4. When is it equal to 4?
we move point P?
point P one more time?
explore why our conjecture that the relationship between these
two triangles is greater than or equal to 4.
triangle DFE is 1/4 of the area of the triangle ABC, so we have
a ratio of 4 to 1.
the ratio equal 4 exactly? Use the link and explore the relationship
when P is on the centriod.