Part A
Consider any 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.
Explore (AF)(BD(EC) and (FB)(DC)(EA) for various locations of P.
Position 1
Using the same triangle, we have (AF)(BD)(EC) = 1.67 inches cubed.
We also have (FB)(AE)(DC)=1.66 inches cubed.
Using a more effect method of measuring the lengths, it is clear that the product of the lengths, (AF)(BD)(ED) is the same as the product of the lengths (FB)(DC)(EA).
POSITION 2
I changed the location of point P. Notice that the product of the lengths, (AF)(BD)(ED) is the same as the product of the lengths (FB)(DC)(EA).
Positions 3
Part B
Conjecture? Prove it!
Theorem:
Given any 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. It follows that [ (AB)(BD)(CE) ] / (BF)(CD)(AE) = 1.
Proof:
To procuce simiar triangles, I constructed parallel lines.
Now, I will prove triangle AFZ is similar to triangle BFP.
I have angle AFZ is equal to angle BFP because verticle angles are equal.
Angle FAZ equals angle FBP because we have two parallel lines cut by a transversal, so alternate interior angles are equal. Similarly, angle AZF equals angle BPF. Since the angles are equivalent, we have triange FBP is similar to triangle FAZ.
Similarly,
Triangle EAT is similar to triangle ECP. Triangle DCP is similar to triangle DBS. Triangle ECM is similar to triangle EAP. Triangle FAP is similar to triangle FBN. Triangle DCY is similar to triangle DBP.
We also can prove that triangle AFP is similar to triangle ABS. Because the two triangles share an angle, we have FAP equals angle BAS. Angle AFP equals angle ABP because we have two parallel lines cut by a transversal so corresponding angles are equal. Similarly angle APF equals angle ASB. Therefore triangle AFP is similar to triangle ABS and triangle BFN.
Similarly,
Triangle EAT is similar to triangle ECP and triangle ACZ.
Triangle DCP is similar to triangle DBS and triangle BCN.
Triangle ECM is similar to triangle EAP and triangle CAY
Triangle DCY is similar to triangle DBP and triangle CBN.
Triangle FBP is similar to triangle FAZ and triangle ABT.
So (FA / FB) = (AZ / BP)
and (CY / CD) (FA / FB) = (AZ / BP) (BP / BD).
Simplying we have (CY / CD) (FA / FB) = (AZ / BD).
Now, (CE / EP) (CY / CD) (FA / FB) = (AZ / BD) (AC / AZ).
Simplying we have (CE / EP) (CY / CD) (FA / FB) = (AC / BD).
Now, (EP / EA) (CE / EP) (CY / CD) (FA / FB) = (AC / BD) (CY / CA).
Simplying we have (CE / EA) (CY / CD) (FA / FB) = (CY / BD)
Multiply the equation by (BD / CY).
Therefore we have (AF)(BD)(CE) / (BF)(CD)(AE) = 1.
Part C
Show that when P is inside triangle ABC, the ratio of the areas of ABC and DEF is always greater than or equal to 4. When is it equal to 4.
Notice that when P is the centroid the ratio is equal to 4.00. Also notice that triangle DEF is the medial.
