Every triangle is equivalent by dissection to a rectangle.

**Hypothesis: **Let $\Delta$ABC be a triangle.

**Construction: ** Suppose
that $\angle$A and $\angle$B are acute angles in the triangle
$\Delta$ABC (Re-label the triangle if needed). Let M be the mid-point
of $\overline{AC}$ and N be the mid-point of $\overline{BC}$. Draw the
line $\overleftrightarrow{MN}$ through M and N. Let D and E be the foot
of the perpendiculars of A and B on the line $\overleftrightarrow{MN}$,
respectively; and let F be the foot of the perpendicular of C on the
line $\overleftrightarrow{MN}$.

From the figure, it is clear
that $\angle$DMA and $\angle$FMC area a pair of vertical angles. And,
$\angle$CNF and $\angle$BNE are another pair of vertical angles.

**To Prove:** Area of $\square$ABED = Area of $\Delta$ABC.

**Proof:**
S. No. | Statement | Reason |

1 | ∠DMA≅∠FMC | By Vertical Angle Theorem |

2 | ∠MDA≅∠MFC | Right angles |

3 | AM=CM | M is the mid-point of $\overline{AC}$ |

4 | ΔADM≅ΔCFM | By AAS property of congruency |

5 | Area of ΔADM = Area of ΔCFM | Areas of congruence triangles |

6 | ∠CNF≅∠BNE | By Vertical Angle Theorem |

7 | ∠CFN≅∠BEN | Right angles |

8 | CN=BN | N is the mid-point of $\overline{BC}$ |

9 | ΔCFN≅ΔBEN | By AAS property of congruency |

10 | Area of ΔCFN = Area of ΔBEN | Areas of congruence triangles |

11 | Area of ▭ABED = Area of ΔBEN+Area of ▭ABNM+Area of ΔADM | Area of whole is equal to sum of the areas of parts |

12 | Area of ▭ABED = Area of ΔCFN+Area of ▭ABNM+Area of ΔCFM | Using (5) and (10) in (11) |

13 | Area of ΔABC= Area of ΔCFN+Area of ▭ABNM+Area of ΔCFM | Area of whole is equal to sum of the areas of parts |

14 | Area of ▭ABED = Area of ΔABC | By (12) and (13) |

Hence, we proved the statement.