Question

Explain theorem 7.1 maths ncert class 9th.


Refer to the given attachment.​

Answer 1

Given:

In ΔABC & ΔDEF

∠B = ∠E

∠C = ∠F

BC = EF

To Prove:

ΔABC≅ΔDEF by ASA congruency rule

Proof:

1) Case 1:

Let AB = DE

In ΔABC & ΔDEF

AB = DE ( assumed)

∠B = ∠E (given)

BC = DE (given)

so, ΔABC ≅ ΔDEF ( By SAS rule)

2) Case 2:

AB > DE

• Construction: take a point M on AB such that MB = DE

In ΔPBC & ΔDEF

MB = DE ( assumed)

∠B = ∠E (given)

BC = DE (given)

so, ΔMBC ≅ ΔDEF ( By SAS rule)

=> ∠MCB = ∠DFE

• But it is given that the ∠ACB = ∠DFE

so ∠ACB = ∠MCB

• This is possible only when point M coincides with A

=> AB = DE

∴ by case 1

ΔABC ≅ ΔDEF

3) Case 3:

AB < DE

• Construction: Take a point N on DE such that EN = AB
• Now follow the case 2 again for AB < DE

• hence we get ΔABC ≅ ΔDEF

In all the cases the triangles are congruent to each other.

Answer 2

ASA Congruence Rule ( Angle – Side – Angle )

Two triangles are said to be congruent if two angles and the included side of one triangle are equal to two angles and the included side of another triangle.

Proof :

From the given two triangles, ABC and DEF in which:

∠B = ∠E, and ∠C = ∠F and the BC = EF

To prove that ∆ ABC ≅ ∆ DEF

For proving congruence of the two triangles, the three cases involved are

Case (i): Let AB = DE

You will observe that

AB = DE (Assumed)

Given ∠B = ∠E and BC = EF

So, from SAS Rule we get, ∆ ABC ≅ ∆ DEF

Case (ii): Let it possible the side AB > DE. Now take a point P on AB such that it becomes

PB = DE.

ASA Congruence Rule proof

Now consider ∆ PBC and ∆ DEF,

IT is noted that in triangle PBC and triangle DEF,

From construction, PB = DE

Given,∠ B = ∠ E

BC = EF

So, we conclude that, from the SAS congruence axiom

∆ PBC ≅ ∆ DEF

Since the triangles are congruent, their corresponding parts of the triangles are also equal.

So, ∠PCB = ∠DFE

But, we are provided with that

∠ACB = ∠DFE

So, we can say ∠ACB = ∠PCB

Is this condition possible?

This condition is possible only if P coincides with A or when BA = ED

So, ∆ ABC ≅ ∆ DEF (From SAS axiom)

Case (iii): If AB < DE, we can take a point M on DE such that it becomes ME = AB and

repeating the arguments as given in Case (ii), we can conclude that AB = DE and so we get

∆ ABC ≅ ∆ DEF.

Suppose now consider that in two triangles, two pairs of angles and one pair of corresponding

sides are equal but the side of a triangle is not included between the corresponding equal pairs of angles. Can you say that the triangles still congruent? Absolutely, You will notice that they are congruent. Because the sum of the three angles of a triangle is 180°. If two pairs of

angles are equal, the third pair of angles are also equal. It is called as AAS congruence rule when two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal