ABCD is a parallelogram. Given that,
DC = CE
prove that,
area of the ADE triangle is equal to the area of the parallelogram.

need explanation

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Drop a perpendicular from to $\overline {DE}$.

Let $l(AK)=h$ and $l(AB)=l(DC)=L$

$h$ is the height of the parallelogram and the triangle.

$L$ is the length of the base of the triangle and one of the sides of parallelogram.

We have area of parallelogram $=A_{para}=hL$

And the area of triangle is:

$A_{tria}=\frac{1}{2} h\times (L+L)=hL$

Thus we can see that they are equal.

See below

Step-by-step explanation:

Given:

ABCD is a parallelogram

DC = CE

To Prove:

area of the ADE triangle is equal to the area of the parallelogram.

Proof:

Let The Point between B and C be F

To prove that area of the ADE triangle is equal to the area of the parallelogram, we'll first prove that ΔADF ≅ ΔECF

Statements                             |               Reason

DC ≅ CE                                  |                Given

But,  CD ≅ AB                         |    Opposite sides of a ║gm

So,  CE ≅ AB                           |   Transitive Property of Equality

∵  ΔABF ↔ ΔECF                    |

 CE ≅ AB                                |       Already Proved

  ∠CFE ≅ ∠ BFE                     |       Vertical Angles

  ∠ECF ≅ ∠ABF                      |      Alternate Angles

So, ΔABF ≅ ΔECF                   |    S.A.A.  Postulate

Now,                                         |

Area of ║gm = Area of            |

quad ADCF + ΔABF                |

But,    ΔABF ≅ ΔECF               |        Already Proved

So,                                            |

Area of ║gm = Area of            |

quad ABCF + ΔECF                 |

But Area of ΔADE = Area of   |

quad ABCF + ΔECF                 |

So,                                            |

Area of ║gm = Area of ΔADE |     Hence Proved

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step-by-step explanation: