Theorem I :
For any triangle ABC, if D and E are on AB and AC respectively, then DE || BC and DE = 1/2 BC iff D, E are midpoints of AB and AC respectively.
To proof this theorem,
Comparing triangles AEF and CDF,
<EAF = <FCD (alternate interior angles as AB || CD)
AF = FC (F is the mid point of AC)
<AFE = <CFD (Vertically opposite angles)
Therefore, triangle AEF is similar to CDF (ASA)
So, EF = DF and AE = DC
Therefore, BE = AE = DC
Hence, BCDE is a parallelogram. (BE = DC and BE || DC)
This gives EF || BC.
Also, EF = DF and EF + DF = ED = BC
Therefore, 2EF = BC
--> EF = 1/2 BC.
A sample question on the midpoint theorem :
Find x.
From Pythagoras' Theorem, we get that the hypotenuse is 10.
The midpoint theorem implies that x is half of the hypotenuse.
Therefore, x = 10/2 = 5.
For any triangle ABC, if D and E are on AB and AC respectively, then DE || BC and DE = 1/2 BC iff D, E are midpoints of AB and AC respectively.
To proof this theorem,
Comparing triangles AEF and CDF,
<EAF = <FCD (alternate interior angles as AB || CD)
AF = FC (F is the mid point of AC)
<AFE = <CFD (Vertically opposite angles)
Therefore, triangle AEF is similar to CDF (ASA)
So, EF = DF and AE = DC
Therefore, BE = AE = DC
Hence, BCDE is a parallelogram. (BE = DC and BE || DC)
This gives EF || BC.
Also, EF = DF and EF + DF = ED = BC
Therefore, 2EF = BC
--> EF = 1/2 BC.
A sample question on the midpoint theorem :
Find x.
From Pythagoras' Theorem, we get that the hypotenuse is 10.
The midpoint theorem implies that x is half of the hypotenuse.
Therefore, x = 10/2 = 5.