**Is a parallelogram a rectangle? Can you call a rectangle a parallelogram?**

**Solution:**

** **A parallelogram becomes a rectangle when its angles measure 90° each. Therefore a parallelogram need not be a rectangle. But a rectangle is a parallelogram

**Prove that bisectors of two opposite angles of parallelogram are parallel.**

Given: ABCD is a | | gm AE and CF bisect DAB and DCB respectively

To prove: AE | | CF

**Proof: **

**Statement Reason **

ABCD is | |gm given

∴ DC | | AB

∠CDA + ∠DAB = 180° Co-interior angles

∴ ∠DAB = 180° – ∠CDA are supplementary

But ∠DAE = ^{1}/_{2} ∠DAB ∵ AE bisects DAB

∴ ∠DAE = ^{1}/_{2} [180° – ∠CDA]

∠DAE = 90° – 12 [180 – ∠CDA] ……. (1)

In Δle ADE

∠DEA = 180° – (∠DAE + ∠ADE )

= 180° – [90°– 12 ∵ ∠ ADE = ∠CDA + ∠ADE ∠CDA from equation(1)

= 180° – [90°– ^{1}/_{2} ∠CDA + ∠CDA] [∵∠CDA – ^{1}/_{2 }∠CDA = ∠CDA]

= 180° – [90°– ^{1}/_{2} ∠CDA]

= 180° – 90°– ^{1}/_{2} ∠CDA

∠DAE = 90° – ^{1}/_{2} ∠CDA …….. (2)

From (1) and (2)

∠DAE = ∠DEA …….. (3)

Further, ∠ECF = ^{1}/_{2} ∠DCB, But ∠DCB = ∠DAB

∠ECF = ^{1}/_{2} ∠DEA

ECF = ∠DAF But from (3) DAE = DEA

∠ECF = ∠DEA

But these are corresponding angles

∴ AE | | FC

**Prove that if the diagonals of a parallelogram are perpendicular to each other, the parallelogram is a rhombus.**

**Solution:**

**Given: **ABCD is a parallelogram. Diagonals AC and BD intersect at right angles.

**To prove: **ABCD is a rhombus

Proof: | reason |

ABCD is parallelogram | given |

AC ⊥ BD | |

In ΔAOD and ΔCOD, | |

OD is common | |

AO = CO | Diagonals bisect each other |

∠AOD = ∠COD = 90° | given |

∴ΔAOD ≅ ΔCOD | SAS |

∴AD = CD | C.P.C.T |

∴ABCD is a rhombus | adjacent side equal |

**Prove that if the diagonals of a parallelogram are equal then it is a rectangle.**

**Solution:**

** **

**Given: **ABCD is a parallelogram. Diagonals AC and BD are equal.

**To prove: **ABCD is a rectangle. ** **

Proof: | reason |

In ΔDAB and ΔCBA | Opposite sides of a parallelogram. |

DA = CB | |

AB = BA | |

DB =CA | given |

∴ΔDAB ≅ ΔCBA | SSS |

∴∠DAB = ∠CBA | CPCT |

But ∠DAB + ∠CBA = 180° | AD | | BC |

∴∠DAB = ∠CBA = ^{180}/_{2} = 90° | angles |

∴ABCD is a rectangle | an angle is a right angles |

## One thought on “QUADRILATERALS – EXERCISE 4.2.1 – Class 9”

Comments are closed.