Question

prove that each angle of a rectangle is 90° without letting any angle 90°​

Answer 1

Answer:

Let us consider a rectangle ABCD.

We know that the diagonals of a rectangle are equal. Therefore, AC = DB

In ΔABC and Δ DCB,

AB = DC (Opposite sides of a parallelogram)

AC = DB

BC = CB (Common side)

Therefore, ΔABC≅ΔDCB

And,  ∠ABC =  ∠DCB (CPCT)

And,  ∠ABC +  ∠DCB = 180° (Adjacent angles of a parallelogram are supplementary/Co interior angles)

Therefore, ∠ ABC =  ∠DCB = 90°

And,  ∠ABC = ∠ ADC = 90° and  ∠DCB = ∠ BAC = 90° (Opposite angles of a parallelogram are equal.)

Hence, each angle of a rectangle measures 90°

Hence proved.

Step-by-step explanation:

Answer 2

Solution:

Note: Diagram of this question attached in attachment file.

In Figure, the diagonals AC and BD intersect at O.

Thus the vertical angles,

∠AOD = ∠BOC

Since the line segments AB and CD are parallel lines, and AC is the transversal, then the alternate interior angles

∠OCB = ∠OAD

Since the line segments AB and CD are parallel lines, and BD is the transversal, then the alternate interior angles

∠OBC = ∠ODA

The sides AD = BC

Therefore, by Angle-Side-Angle congruence,

ΔAOD = ΔBOC

AB and CD are parallel, and BD is a transversal, therefore the angles,

∠CDB = ∠ABD

Now,

∠ADC = ∠ODA + ∠BDC

and

∠ABC = ∠OBC + ABD

Thus,  ∠ADC = ∠ABC

We also know that in a parallelogram, the sum of opposite angles are 180 degrees.

Also,

∠ADC + ∠ABC = 180°

So,

∠ADC = ∠ABC

They both have equal to 90°.

Thus, in a rectangle all the angles are 90 degrees.