Prove that Diagonal of a parallelogram divides it into two congruent t

1.	Prove that Diagonal of a parallelogram divides it into two congruent triangle?
Answer» Step-by-step explanation: In parallelogram ABCD, Angle A = Angle C .......(since opp. ANGLES of a llgm are equal) .....1 Similarly , Angle B = Angle D Construct a diagnal BD Now , angle ABD = angle BDC ...(diagnols bisect the angles of a llgm).....2 Side AD = side BC....(opp. sides of llgm are equal)..3 From 1, 2, 3 By ASA CONGRUENCY rule triangle ABD = triangle BDC . *HENCE* *PROVED*

Prove that Diagonal of a parallelogram divides it into two congruent triangle?

Answer»

Step-by-step explanation:

In parallelogram ABCD,

Angle A = Angle C .......(since opp. ANGLES of a llgm are equal) .....1

Similarly , Angle B = Angle D

Construct a diagnal BD

Now , angle ABD = angle BDC ...(diagnols bisect the angles of a llgm).....2

Side AD = side BC....(opp. sides of llgm are equal)..3

From 1, 2, 3

By ASA CONGRUENCY rule

triangle ABD = triangle BDC

. HENCE PROVED

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Prove that Diagonal of a parallelogram divides it into two congruent triangle?​

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Prove that Diagonal of a parallelogram divides it into two congruent triangle?