1. TBQ - Prove the following I

 

 

 

Proof:  

We know that the rectangle is a rhombusparallelogramsquare in which one angle is a right angle.

 

Let \(ABCD\) be a rectangle in which \(∠ A = 90^°\).
 
To Prove: \(∠ B = ∠ C = ∠ D = 90^°\)
 

 We have, \(AD || BC\) and \(AB\) is a transversal.
 
So, \(∠ A + ∠ B = 180^°\) [linear pair of anglesInterior angles on the same side of the transversalcorresponding angles]
 
But, \(∠ A = 90^°\)
 
So, \(∠ B = 180^° – \)\(\angle A\)\(\angle B\)\(\angle C\)\(= 180^° – 90^° =\)\(180^\circ\)\(360^\circ\)\(90^\circ\)
 
Now, \(∠ C =\) \(\angle A\)\(\angle B\)\(\angle C\) and \(∠ D =\) \(\angle C\)\(\angle A\)\(\angle B\) (Opposite angles of the parallellogram)
 
So, \(∠ C = 90^°\) and \(∠ D = 90^°\).
 

Therefore, each of the angles of a rectangle is a acute angle right angleparallel.