Square and its Theorems

In this section we will discuss square and its theorems.
A square is a parallelogram with all sides equal and all angles are 90
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Square and its Theorems :

Theorem 1 : The diagonals of a square are equal and perpendicular to each other. Given : ABCD is a square.

Prove that : AC = BD and AC ⊥ BD .

 Statements Reasons 1) ABCD is a square. 1) Given 2) AD = BC 2) Properties of square. 3) ∠BAD = ∠ABC 3) Each 900 and by properties of square. 4) AB = BA 4) Reflexive (common side) 5) Δ ADB ≅ ΔBCA 5) SAS postulate 6) AC = BD 6) CPCTC 7) OB = OD 7) As square is a parallelogram so diagonals of parallelogram bisect each other. 8) AB = AD 8) Properties of square. 9) AO = AO 9) Reflexive (common side) 10) ΔAOB ≅ ΔAOD 10) SSS Postulate 11)∠AOB = ∠AOD 11) CPCTC 12) ∠AOB + ∠AOD = 180 12) These two angles form linear pair and Linear pair angles are supplementary). 13) 2∠AOB = 180 13) Addition property. 14) ∠AOB = 90 14) Division property. 15) AO ⊥ BD ⇒ AC ⊥ BD 15) Definition of perpendicular.
Theorem 2 : If the diagonals of a parallelogram are equal and intersect at right angles, then the parallelogram is a square. Given : ABCD is parallelogram in which AC = BD and AC ⊥ BD.

Prove that : ABCD is a square.

 Statements Reasons 1) ABCD is a parallelogram 1) Given 2) AC = BD and AC ⊥ BD 2) Given 3) AO = AO 3) Reflexive 4) ∠AOB = ∠AOD 4) Each 900 5) OB = OD 5) Properties of parallelogram. 6) ΔAOB ≅ ΔAOD 6) SAS Postulate 7) AB = AD 7) CPCTC 8) AB = CD and AD = BC 8) Properties of parallelogram. 9) AB = BC = CD = AD 9) From above 10) AB = AB 10) Reflexive (common side) 11) AD = BC 11) Properties of parallelogram. 12) AC = BD 12) Given 13) ΔABD ≅ Δ BAC 13) SSS Postulate 14) ∠DAB = ∠CBA 14) CPCTC 15)∠DAB + ∠CBA = 180 15) Interior angles on the same side of the transversal. 16) 2∠DAB = 180 16) Addition property 17) ∠DAB = ∠CBA = 90 17) Division property

Practice

Here is a square drawn for you. Answer the following questions on the basis of square and its theorems (
m ---> measure ). a. (i) m∠A = ------- (ii) m∠B = -------- (iii) m∠C = -------

b. (i) seg(AB) = ------- (ii) seg (BC) = -------- (iii) seg (CD) = -------

C. (i) seg(AC) = ------- (ii) seg (BD) = -------- (iii) seg (BO) = -------

d. (i) seg(AO) = ------- (ii) seg (CO) = --------

e. (i)m∠DOA = ------ (ii) m∠AOB = ------ (iii) m∠BOC = ------

f. (i) Is AB || CD (ii) Is BC || AD