Geometry Proof 2

In this section we will discuss GeometryProof 2 based on mid point theorem in triangle. If the line joins the mid point of any two sides of a triangle then it is parallel to third side and it one half of the third side.

1) In a triangle ABC, P is the mid point of AB, such that PQ || BC.
Given : 1) P is the mid point of AB ⇒ AP = PB.
2) PQ || BC
Prove that : AQ = QC ⇒ Q is the mid point of AC.
Given : ΔABC, P is the mid point of AB and
PQ ||BC
Prove that : Q is the mid point of AC.
Statements
Reasons
1) P is the mid point 1) Given (ΔABC)
2) AP = PB 2) Definition of mid point
3) AP / PB = 1 3) Ratio of two equal side is 1.
4) PQ || BC 4) Given (ΔABC)
5) ( AP / PB ) = ( AQ / QC ) 5) Basic Proportionality Theorem
6) AQ / QC = 1 6) From (3) and (5) (Transitivity)
7) AQ = QC 7) Cross multiply
8) Q is the mid point of AC 8) Definition of mid point

-------------------------------------------------------------------------
2) ABCD is a trapezoid in which AB || DC and its diagonals intersect each other at 'O'.
Given : 1) ABCD is a trapezoid.
2) AB || DC
Prove that : OE || DC.
Construction : Draw OE such that OE ||DC
Statements
Reasons
1) OE || DC 1) Construction (ΔABC)
2) (BO / OD) = ( BE / EC) 2) Basic Proportionality Theorem
3) AB || DC 3) Given
4) OE || AB 4) From (1) and (3) (Transitivity)
5) (AO / CO) = (BE / EC) 5) Basic Proportionality Theorem
6) (AO / CO) = (BO / DO) 6) From (2) and (5) ( Transitivity)
7) (AO / BO) = (CO / DO) 7) From above ( Alternendo)

• Geometry proofs
• GeometryProof-1
• GeometryProof 2
• Proofs on Area of similar triangles
• Pythagorean theorem

Home