The diagonal of a parallelogram divides the parallelogram into two congruent triangles prove it

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Diagonal AC divides the parallelogram into two triangles $\vartriangle $ABC and $\vartriangle $ADC.In $\vartriangle $ABC and $\vartriangle $ADC:$\because $ AD||BC$\angle $BAC = $\angle $DCA ( By alternate angle)AC = AC (Common side)$\angle $BCA = $\angle $DAC ( By alternate angle)In these two triangles, one side and two angles made on this side are equal.Therefore by ASA rule of congruence: $\vartriangle $ABC $ \cong $ $\vartriangle $ADC.Since, both these triangles are congruent. So, all the corresponding sides and angles of one triangle are equal to that of the other.$\therefore $ AD= BC And AB = CD.Therefore, it is proved that the diagonal of a parallelogram divides it into two congruent triangles and also opposite sides of a parallelogram are equal.Note- In the question where you have to show two triangles congruent. You should remember the following rule of congruence:1.SSS (All corresponding sides of one triangle is equal to other triangle)2.SAS (Two sides and one angle between the two sides of one triangle is equal to the other)3. ASA (one side and two angles made on this side are equal)4.RHS ( This is for right triangles. One side and hypotenuse of one right triangle is equal to other right triangle)

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A diagonal of a parallelogram divides it into two congruent triangles.

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