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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

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