1.

Show that the diagonals of a square are equal and bisect each other at right angles.

Answer»

Data: ABCD is a parallelogram. Its diagonals are AC and BD. They meet at ‘O’. 

To Prove: i) AO = OC BO = OD 

ii) ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°. 

Proof: In ∆ABC and ∆ABD, 

BC = AD (diagonals of square is equal) 

∠ABC = ∠BAD = 90° 

(angles of square) 

AB is common. 

∴ ∆ABC ≅ ∆ABD (SAS Postulate.) 

In ∆AOB and ∆COD, 

AB = DC (sides of square) 

∠OAB = ∠OCD (alternate angles) 

∠OBA = ∠ODC (alternate angles) 

∴ ∆AOB ≅ ∆COD (ASA Postulate) 

AO = OC 

BO = OD …………….. (i) 

Similarly, ∆AOB ≅ ∆BOC. 

∴ ∠AOB = ∠BOC = 90° 

Now, ∆COD ≅ ∆AOD, then 

∴ ∠COD = ∠DOA = 90° 

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90° …………. (ii) 

from (i) and (ii) 

∴ Sides of a square are equal and bisect at right angles.


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