Question 4

Consider any triangle ABC. Select a point inside the triangle and draw lines AP, BP and CP extended to meet the opposite sides in D, E and F respectively.


Claim (1):

For all positions of P inside triangle ABC the product (AF)(BD)(AE) will always equal the product (FB)(DC)(EA).

For a GSP animation that illustrates this claim please click here.

To help us in proving the claim we will consider the ratio:

Or more particularly the product of ratios:

Proof of claim (1):

We begin by constructing a line through A that is parallel to BC and meets lines CF and BE in S and T respectively.

The construction creates a number of similar triangles.

By the pairs of similar triangles SAP and CDP and TAP and BDP:

We can conclude that:
By the pairs of similar triangles TAE and BCE and SAF and CBF:

We can conclude that:
After multiplying we note that:

From whence it follows that (AF)(BD)(AE) = (FB)(DC)(EA)!

NOTE the basis of the proof is a theorem known as Ceva's Theorem -- which we discussed in class during the quarter.

Claim (2):

For all positions of P outside the triangle ABC the product (AF)(BD)(AE) will always equal the product (FB)(DC)(EA).

NOTE:
Clearly the case when P lies on triangle ABC is a trivial case as one of the dimensions in each product goes to zero!

For a GSP animation that illustrates claim (2) please click here.

To help us in proving the claim we will again consider the ratios:

Proof of claim (2):

Please note while this is technically the proof of one case - that is you could argue that there are different possible positions for P, it is left to the reader to consider the fact that this proof generalises to all positions of P.

We begin by constructing a line through A that is parallel to BC and meets lines CF and BE in T and S respectively.

The construction creates a number of similar triangles.

By considering the pairs of similar triangles SAP and BDP and TAP and CDP:

We can conclude that:
By considering the similar triangles ATF and BCF:

We can conclude that:
By considering the similar triangles ASE and CBE:

We can conclude that:

After multiplying we note that:

From whence it follows that (AF)(BD)(AE) = (FB)(DC)(EA)!

Claim (3):

For all positions of P inside triangle ABC the ratio (area ABC) to (area FED) will always be greater than or equal to 4.

For a GSP animation that illustrates this claim please click here.

I will not prove the claim in full but rather provide an outline of the approach.

Clearly we can make the area of triangle FED very small, indeed it approaches zero as P approaches the triangle ABC (refer to previous remarks).

Since we know that the area of the medial triangle is one quarter of the area of the parent triangle we know that the ratio will equal 4 when P is the Centroid (G).

What remains to show is that the triangle FED cannot be greater than one quarter of the area of triangle ABC.

Back