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|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)|
|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|