Question

prove that diagonal of a parallelogram bisect each other.​

Answer 1

Answer:

prooved

Step-by-step explanation:

in attachment

Answer 2

$\huge\mathcal\purple{Answer:}$

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

A parallelogram ABCD such that its diagonals AC and BD intersect at O.

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

OA = OC and OB = OD

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

Since ABCD is a llgm. Therefore,

AB || DC and AD || BC.

➜Now, AB || DC and transversal AC intersect them at A and C respectively.

$\therefore$ $\angle$BAC = DCA. _________ ( °•° Alternate interior angles are equal).

$\implies$ $\angle$BAO = DCO. ___(i)

___________________________

➜Again, AB || DC and BD intersects them at B and D respectively.

$\therefore$ $\angle$ABD = $\angle$CDA ______________ (°•° Alternate interior angles are equal)

$\implies$ $\angle$ABO = $\angle$CDO __________ (II)

______________

Now, in ∆s ABO and COD, we have

1. $\angle$BAO = DCO____(from (i)
2. BA = CD ______ (opposite sides of a llgm are equal)
3. $\angle$ABO = $\angle$CDO ____ (from (ii)

So, by ASA congruence criterion

∆AOB $\cong$ ∆COD.

= OA = OC and OB = OD

HENCE, OA = OC AND OB = OD.