Question

if the diagonal of a parallelogram are equall then show that it is a rectangle

Answer 1

let ABCD a  ||gm
: opposite sides eaqual AB=CD and AD=BC
AC=BD |given
AB=BA |common
Therfore triangle ACB is Congurrent to triangle BDA
<ACB = < BAD (1) | by cpct
Again AD || BC
AD || BC and transversal AB intersects thm
<BAD + <ABC =180 (2)| SUm of consecutive interrior angleson the same side of the transversal is 180 degree

FROM (1) and (2)
<BAD=<ABC=90 degree
therfore <A= 90 degree and <b =90 degree
Similarly we can prove <c=90 degree and <D=90 degree

Therfore ABCD is A triangle

NOTE: < is used for ANGLE and ||gm for parallelogram and || for parallel

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Answer 2

Let the parallelogram be ABCD and AC and BD are its diagonals
In ΔADC and ΔBCD 
AC=BD(given)
AD=BC(opp. sides of a parallelogram are equal)
CD=CD(common)
∴By SSS congruence rule 
ΔADC is congruent to ΔBCD
by cpct
∠ADC=∠BCD
Since in a parallelogram adjacent angles are supplementary
∠ADC+∠BCD=180°
2∠ADC=180°(∠ADC=∠BCD)
∴∠ADC=90°
and since ∠ADC=∠BCD
∠BCD=90°
since in a parallelogram adjacent angles are 90°, the parallelogram is a rectangle.
Hence Proved