PART ONE

PART TWO

A. Consider any triangle ABC. Select any point P inside triangle ABC and draw lines AP, BP, CPextended to their intersections with the opposite sides in points D, E, and F respectively.

Explore (AF) (BD) (CE) and (FB) (DC) (EA) for various triangles and various positions of P.

This reminds of Assignment 8 - Problem 12. Therefore, I will use an animation file that includes all triangle types (acute, right, obtuse) with P either inside, or outside.

Rather than separately exploring (AF) (BD) (CE) and (FB) (DC) (EA) , I want to explore the ratio (this is similar to what I did in Assignment 8)

CASE ONE: P IS INSIDE ABC

For any triangle type, we get (AF) (BD) (CE) = (FB) (DC) (EA). Click the figure below to see this.

CASE TWO: P IS OUTSIDE ABC

For any triangle type, we get (AF) (BD) (CE) = (FB) (DC) (EA). Click the figure below to see this.

B. Prove that (AF) (BD) (CE) = (FB) (DC) (EA) for any point P on the plane

I will imitate the area modeling I used in Assignment 8 - Problem 12 with similarity theorems as well.

First let's see that the area of ABC is equal to the sum of the areas of APB, BPC, and APC

Area Triangle APB + Area Triangle BPC + Area Triangle APC = Area Triangle ABC

Now it's a good idea to write the areas of each triangle in terms of theirs bases and heights: AREA = 0.5 x BASE x HEIGHT. We know their bases, but we don't know their heights. Let's define altitudes corresponding to each triangle:

Area modeling once again. See figure below:

Express the ratio BD / DC as the ratio of the areas of two triangles: Triangles PBD and PDC have the same altitudes PR. Therefore, the ratio of their areas equals the ratio of the lengths of the sides to which these altitudes belong. In other words, Area(PBD) / Area(PDC) = BD / DC.

Similarly , express the ratio CE / EAas the ratio of the areas of two triangles: Triangles PCE and PEA have the same altitudes PS. Therefore, the ratio of their areas equals the ratio of the lengths of the sides to which these altitudes belong. In other words, Area(PCE) / Area(PEA) = CE / EA.

Similarly the ratio AF / FB as the ratio of the areas of two triangles: Triangles PAF and PFB have the same altitudes PT. Therefore, the ratio of their areas equals the ratio of the lengths of the sides to which these altitudes belong. In other words, Area(PAF) / Area(PFB) = AF / FB.

Therefore, (BD / DC)( CE / EA)(AF / FB) = [Area(PBD) / Area(PDC)] [Area(PCE) / Area(PEA)] [Area(PAF) / Area(PFB)]

I could not cancel the right hand side.

Express the ratio BD / DC as the ratio of the areas of two triangles: Triangles ABD and ADC have the same altitudes AK. Therefore, the ratio of their areas equals the ratio of the lengths of the sides to which these altitudes belong. In other words, Area(ABD) / Area(ADC) = BD / DC.

Similarly , express the ratio CE / EAas the ratio of the areas of two triangles: Triangles BCE and BEA have the same altitudes BL. Therefore, the ratio of their areas equals the ratio of the lengths of the sides to which these altitudes belong.as the ratio of their bases. In other words, Area(BCE) / Area(BEA) = CE / EA.

Similarly the ratio AF / FB as the ratio of the areas of two triangles: Triangles CAF and CFB have the same altitudes CM. Therefore, the ratio of their areas equals the ratio of the lengths of the sides to which these altitudes belong.as the ratio of their bases. In other words, Area(CAF) / Area(CFB) = AF / FB.

Therefore, (BD / DC)( CE / EA)(AF / FB) = [Area(ABD) / Area(ADC)] [Area(BCE) / Area(BEA)] [Area(CAF) / Area(CFB)]

Once again I could not cancel the right hand side.

I give up on area modeling and go back to my high school years. I want to use the basic similarity theorems I learned in high school.

Consider the extension of one of the sides of ABC, say BC. Let a line d intersect two sides AB and AC of the triangle ABC and the extension of BC at points Q, R, S, respectively.

Now let us draw the line k parallel to side AC and passing through point B. Let T be the intersection point of lines k and d.

Use Angle-Angle-Angle Similarity Theorem: QBT ~ QCS

From this, we get ( QB / QC ) = ( BT / CS ) -----> EQUATION ONE

Furthermore, we get BRT ~ ARS (By Thales Theorem)

From this, we get ( BR / AR) = ( BT / AS ) -----> EQUATION TWO

Dividing EQUATION ONE and EQUATION TWO side by side gives

( QB / QC ) ( AR / BR) = ( AS / CS )

Dividing both sides of the equality by ( AS / CS ) gives

( QB / QC ) ( AR / BR) ( CS / AS ) = 1-------> EQUATION THREE

Now it remains to use this EQUATION THREE in our original figure a couple of times:

STEP ONE: Apply EQUATION THREE in triangle ADC with respect to line that passes through points B and E:

( BD / BC ) ( CE / EA) ( AP / PD) = 1----------> EQUATION FOUR

STEP TWO: Apply EQUATION THREE in triangle ABD with respect to line that passes through points Cand F:

( DC / BC ) ( FB / AF) ( AP / PD) = 1----------> EQUATION FIVE

If we divide EQUATION FOUR and EQUATION FIVE side by side, we get:

( BD / DC ) ( CE / EA) ( AF / FB) = 1----------> EQUATION SIX

This completes the proof.

C. Show that when P is inside triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4. When is it equal to 4?