Question

Prove that a diagonal of a parallelogram divides it into two congruent triangles. Also, show that the line
segments joining the mid ponits of the opposite sides of a quadrilateral bisect each other.​

Answer 1

Step-by-step explanation:

Given : ABCD is a parallelogram and AC is diagonal.

To prove :

Triangle ABC ~= Triangle ADC

Proof : Opposite sides of parallelogram is parallel.

So , AB || DC & AD || BC

Since , AB || DC and AC is transversal

Angle BAC = Angle DCA (alternate angles) ----- (i)

Since , AD || BC and AC is transversal

Angle DAC = Angle BCA (alternate angles) -----(ii)

In triangle ABC and ADC ,

Angle BAC = Angle DCA (by alternate property)

AC = AC ( Common )

Angle DAC = Angle BCA (by alternate property)

By ASA Rule ,

Triangle ABC ~= Triangle ADC

Hence proved ..

Answer 2

Let us consider ABCD be a parallelogram in which E and F are mid-points of AB and CD. Join EF.

To prove: ar (|| AEFD) = ar (|| EBCF)

Let us construct DG ⊥ AG and let DG = h where, h is the altitude on side AB.

Proof:

ar (|| ABCD) = AB × h

ar (|| AEFD) = AE × h

= ½ AB × h ….. (1) [Since, E is the mid-point of AB]

ar (|| EBCF) = EF × h

= ½ AB × h …… (2) [Since, E is the mid-point of AB]

From (1) and (2)

ar (|| ABFD) = ar (|| EBCF)

Hence proved.

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