Question

proof that diagonal of parellelogram are equal and bisect each other at right angle, then it is square.

Answer 1

Answer:

एनडीजेडीजेडीएचजेजिध

Step-by-step explanation:

एनएसएनएक्सआईबीडीजेएक्संजजा

Answer 2

Answer:

Step-by-step explanation:

let a quadrilateral be ABCD and its diagonals intersect at point O forming right angles.since the diagonals are equal all the four bisected segments are also equal.

∴ OA = OB =OC =OD ---------- 1

now in the following figure, in ΔAOB and ΔAOD

AO=AO            ----------  common side

∠AOB= ∠AOD ----------   each equal to 90°

OB =OD          -----------   from 1

∴ by S-A-S criterion of congurency

ΔAOB  is congruent to Δ AOD

∴ AB=AD ---------- c.p.c.t

similarly  

AB = BC

BC =CD  

CD = AD  

finally, AB = BC = CD =AD --------------- 2

In ΔAOB  

OA=OB

∴∠OAB=∠OBA-----------angles opposite to equal sides are equal

 

Now,  

∠AOB+∠OAB+∠OBA=180°----------------- ANGLE SUM PROPERTY OF Δ

90° + 2∠OAB = 180°

∴∠OAB = ∠OBA = 45 °

similarly,  

∠ OBC = ∠OCB = 45°

∠OCD = ∠ODC = 45°

∠ODA = ∠OAD = 45°

Now,  

∠ABC=∠OBA + ∠OBC

         = 45°+45°

         =90°

similarly

∠BCD =90°

∠CDA = 90°

∠DAB = 90°

 

Now ,

∠ABC=∠BCD=∠CDA=∠DAB=90°-------------------3

 

From 2 and 3 all the angles are right angles and all the sides are equal  

∴ Quadrilateral ABCD is a SQUARE.