# Mid Point Theorem

Some solved problems on Mid Point -Theorem

1) In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Prove that the line segments AF and EC trisect the diagonal BD.

Given : ABCD is a parallelogram. E and F are mid points.

Prove that : AF and EC trisects the diagonal BD i.e BP = PQ = QD

 Statements Reasons 1) E and F are mid points of AB and CD respectively 1) Given 2) AE = 1/2 AB and CF = 1/2 CD 2) Definition of mid point 3) ABCD is a parallelogram 3) Given 4)AB =CD and AB || CD 4) Properties of parallelogram 5) AE = FC and AE || FC 5) From (2) and (4) 6) AECF is a parallelogram 6) Properties of parallelogram and from (5) 7) FA || CE and FQ || CP 7) Properties of Parallelogram 8) F is the mid point of CD and FQ ||CP 8) By Mid Point -Theorem 9) Q is the mid point of DP ⇒PQ = QD 9) By Mid Point -Theorem 10)Similarly, P is the mid point of BQ ⇒ BP = PQ 10) By mid point theorem 11) BP= PQ = QD 11) From (9) and (10)
AF and EC trisects the diagonal BD.

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2) In ΔABC , right angled at B; and P is the mid point of AC. Prove that 1) PQ ⊥ AB 2) Q is the mid point of AB 3) PB = PA = ½ AC

Given : ΔABC right angled at B
P is the mid point of AC

Prove that : 1) 1) PQ ⊥ AB 2) Q is the mid point of AB 3) PB = PA = ½ AC

Construction : Through P draw PQ || BC meeting AB at Q.

 Statements Reasons 1) PQ || BC 1) Given 2) ∠AQP = ∠ABC 2) Corresponding angles 3) ∠AQP = 900 3) Since ∠ABC =900 4) ∠AQP + ∠BQP = 1800 4) Linear pair angles 5) ∠AQP = ∠BQP = 900 5) ∠AQP = 90 and from (4) 6) PQ ⊥ AB 6) From (5) 7) P is the mid point of AC and PQ ||BC 7) Given 8) Q is the mid point. AQ =BQ 8) By mid point theorem and definition of mid point. 9) ∠AQP = ∠BQP 9) From (5) 10) PQ =PQ 10) Reflexive (common) 11) ΔAPQ = ΔBPQ 11) SAS Postulate 12) PA = PB 12) CPCTC 13) PA = 1/2 AC 13) Since P is the mid point of AC
PA = PB = 1/2 AC