﻿ Class 9 NCERT Math Solution
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TOPICS
Unit-9(Theorems)

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

Solution :-
``` 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.

Solution :-
``` 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.

Solution :-
``` 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.

Solution :-
``` 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)
∠ BAC + ∠ DAC = ∠ DCA + ∠ BCA
∠ DAB = ∠ BCD
Similarly, ∠ ADC = ∠ ABC.
```

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

Solution :-
``` 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.

Solution :-
``` 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.
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.

Solution :-
``` 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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