TOPICS

Unit-9(Theorems)

Quadrilaterals

**Theorem-1 :-** A diagonal of a parallelogram divides it into two congruent triangles.

Given that : ABCD is a Parallelogram and AC be diagonal. Prove that : Δ ABC ≅ Δ CDA . Proof : The diagonal AC divides parallelogram ABCD into two triangles, namely, Δ ABC and Δ CDA. In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. So, ∠ BCA = ∠ DAC (Pair of alternate angles) Also, AB || DC and AC is a transversal. So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) So, Δ ABC ≅ Δ CDA (ASA rule).

**Theorem-2 :-** In a parallelogram, opposite sides are equal.

Given that : ABCD is a Parallelogram and AC be diagonal. Prove that : AB = CD and AD = BC. Proof : The diagonal AC divides parallelogram ABCD into two triangles, namely, Δ ABC and Δ CDA. In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. So, ∠ BCA = ∠ DAC (Pair of alternate angles) Also, AB || DC and AC is a transversal. So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) So, Δ ABC ≅ Δ CDA (ASA rule). By CPCT rule, AB = CD and AD = BC.

**Theorem-3 :-** If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Given that : Let sides AB and CD of the quadrilateral ABCD be equal and also AD = BC. Prove that : ABCD is a parallelogram. Construction : Draw diagonal AC. Proof : In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. So, ∠ BCA = ∠ DAC (Pair of alternate angles) Also, AB || DC and AC is a transversal. So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) So, Δ ABC ≅ Δ CDA (ASA rule). By CPCT rule, AB = CD and AD = BC. So, ∠ BAC = ∠ DCA and ∠ BCA = ∠ DA. Hence, ABCD is a Parallelogram.

**Theorem-4 :-** In a parallelogram, opposite angles are equal.

Given that : ABCD is a Parallelogram. Prove that : ∠ DAB = ∠ BCD and ∠ ADC = ∠ ABC. Construction : Draw AC be diagonal. Proof : The diagonal AC divides parallelogram ABCD into two triangles, namely, Δ ABC and Δ CDA. In Δ ABC and Δ CDA, note that BC || AD and AC is a transversal. So, ∠ BCA = ∠ DAC (Pair of alternate angles) Also, AB || DC and AC is a transversal. So, ∠ BAC = ∠ DCA (Pair of alternate angles) and AC = CA (Common) So, Δ ABC ≅ Δ CDA (ASA rule). By CPCT rule, ∠ BAC = ∠ DCA ....(i) and By CPCT rule, ∠ DAC = ∠ BCA ....(ii) Adding eq(i) and eq(ii) ∠ BAC + ∠ DAC = ∠ DCA + ∠ BCA ∠ DAB = ∠ BCD Similarly, ∠ ADC = ∠ ABC.

**Theorem-5 :-** The diagonals of a parallelogram bisect each other.

Given that : ABCD is a Parallelogram. AC and BD are be diagonal whoose intersect point on O. Prove that : OA = OC and OB = OD. Proof : In Δ AOD and Δ BOC, ∠ BCO = ∠ DAO (Pair of alternate angles) ∠ CBO = ∠ ADO (Pair of alternate angles) and BC = DA (parallelogram sides are equal) So, Δ AOD ≅ Δ BOC (ASA rule). By CPCT rule, OB = OD and OA = OC.

**Theorem-6 :-** : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Given that : ABCD is a Parallelogram. AC and BD are be diagonal whoose intersect point on O. AO = OC, and OB = OD. Prove that : ABCD is a Parallelogram. Proof : In Δ AOD and Δ BOC, AO = OC (Given) OB = OD (Given) and ∠BOC = ∠DOA (parallelogram sides are equal) So, Δ AOD ≅ Δ BOC (SAS rule). By CPCT rule, ∠ BCO = ∠ DAO and ∠ CBO = ∠ ADO. So, AB||CD and BC||AD. Hence, ABCD is a Parallelogram.

**Theorem-7 :-** The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Given that : In figure E and F are mid-points of AB and AC respectively and CD || BA. So, AF = CF. Prove that : EF || BC. Proof : In Δ AEF and Δ CDF ∠ EAF = ∠ DCF (Pair of alternate angles) ∠ AEF = ∠ CDF (Pair of alternate angles) and AF = CF (Given) So, Δ AEF ≅ Δ CDF (ASA rule). By CPCT rule, EF = DF and BE = AE = DC Therefore, BCDE is a parallelogram. This gives EF || BC.

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