Q2) GIVEN: ABCD is a ||gm, where diagonals AC = BD Q4) GIVEN: a square ABCD, whose diagonals AC & BD intersect at O
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Answer:
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1 TO PROVE: ABCD is a rectangle
PROOF: In ∆ADB & ∆ACB
AD = BC (opposite sides of ||gms are equal)
AB = AB (common sides)
BD = AC (given)
Thus ∆ADB ≅ ∆BCA by SSS
hence; ∠A = ∠B by CPCT
Also, ∠A + ∠B = 180° (Adjacent angles are supplementary)
2∠A = 180°
∠A = 90° = ∠B
Thus, ABCD is a ||gm and its one angle is 90° so all angles are 90°
2 TO PROVE: AC = BD, OD=OB, OA=OC; AC⊥BD
A B
D C
A B
D C
O
o
PROOF: In ∆ADB & ∆ACB
AD = BC (opposite sides are equal)
∠A = ∠B = 90° (Angles of square are right angles)
AB = AB (common)
Thus, ∆ADB ≅ ∆BCA by SAS, Thus BD = AC by CPCT
In ∆OAB & ∆ODC:
AB = DC (opposite sides are equal)
∠OAB = ∠OCD (Alternate interior angles are equal)
∠OBA = ∠ODC (Alternate interior angles are equal)
Thus, ∆OAB ≅ ∆OCD by ASA rule; thus OA = OC & OB = OD by CPCT
In ∆AOB & ∆AOD:
AB = AD (Adjacent sides are equal)
OB = OD (just proved above)
AO = AO (common)
Thus, ∆AOB ≅ ∆AOD by SSS rule; thus ∠AOB = ∠AOD by CPCT
∠AOB + ∠AOD = 180° (by linear pair axiom)
2∠AOB = 180°
∠AOB = 90° = ∠AOD
thus AC⊥BD, hence the result
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Step-by-step explanation:
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Answer:
PROOF: In ∆ADB & ∆ACB
AD = BC (opposite sides are equal)
∠A = ∠B = 90° (Angles of square are right angles)
AB = AB (common)
Thus, ∆ADB ≅ ∆BCA by SAS, Thus BD = AC by CPCT
In ∆OAB & ∆ODC:
AB = DC (opposite sides are equal)
∠OAB = ∠OCD (Alternate interior angles are equal)
∠OBA = ∠ODC (Alternate interior angles are equal)
Thus, ∆OAB ≅ ∆OCD by ASA rule; thus OA = OC & OB = OD by CPCT
In ∆AOB & ∆AOD:
AB = AD (Adjacent sides are equal)
OB = OD (just proved above)
AO = AO (common)
Thus, ∆AOB ≅ ∆AOD by SSS rule; thus ∠AOB = ∠AOD by CPCT
∠AOB + ∠AOD = 180° (by linear pair axiom)
2∠AOB = 180°
∠AOB = 90° = ∠AOD
thus AC⊥BD, hence the result
Step-by-step explanation: