Question

a parallelogram has all its sides and angles equal, prove that it is a square.​

Answer 1

Answer:

Bhai, //gm=square

Step-by-step explanation:

Step 1 . first all sides are eq. (given)

square have all sides equal

Step 2. then all angles are equal (given)

So, means sum of all angles of quad.=360°

Then angle1 =360/4=90°

and sq. have its on rang.measured 90°

Step 3. All sides are parallel (given)

Hence proved

Answer 2

${\underline{\underline{\red{\sf{Given:}}}}}$

• There is a paralleogram.
• All sides of paralleogram are equal.
• All angles are equal.

${\underline{\underline{\red{\sf{To\: Prove:}}}}}$

• The given paralleogram is a square.

${\underline{\underline{\red{\sf{Construction:}}}}}$

• Join B to D .
• Join A to C .

${\underline{\underline{\red{\sf{Proof:}}}}}$

Let us first construct a paralleogram ABCD .

(For figure refer to attachment:)

From figure it is clear that AC and BD intersect at O .

Now in ∆AOD & ∆ COD;

1. OD =OD (common)
2. AD =CD (equal sides of paralleogram)
3. AO = CO (diagonals of paralleogramgm bisect each other)

$\orange{\sf{\leadsto \triangle AOD \cong \triangle COD}}$

(By SSS congruence condition.)

So , $\pink{\sf{\angle AOD=\angle COD}}$ (cpct)

Atq ,

$\sf{\implies \angle AOD+\angle COD=180^{\circ}}$ (linear pair)

$\sf{\implies \angle AOD+\angle AOD=180^{\circ}}$

$\sf{\implies 2\angle AOD=180^{\circ}}$

$\sf{\angle AOD=90^{\circ}}$

Hence diagonals bisect at right angles and all angles are equal .

So the given figure is a square.

$\sf{\red{Hence\: Proved}}$