## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

**Particular enunciation: **

In the ∆ABC and ∆DEF, AB = DE, AC = DF and BC = EF,

We have to proved that ∆ABC ≌∆DEF.

**Proof:** Let us assume that BC and EF are the largest sides of the ∆ABC and ∆DEF respectively.

We now apply the ∆ABC to the ∆DEF in such a way that the point B falls on the point E and the side BC falls along the side EF but the point A falls on the side of EF opposite the point D. Let the G be the new position of the point A. Since BC = EF, the point C falls on the point F.

So, ∆GEF is the new position of the ∆ABC.

That is EG = BA. FG = CA and ∠EGF = ∠BAC.

We join D and G. Now, in the ∆EGD, EG = ED [since EG =BA = ED]

Therefore, ∠EDG = ∠EGD

Again, in the ∆FGD, FG = FD, [Since FG = CA = FD]

Therefore, ∠FDG = ∠FGD**, **

So, ∠EDG = ∠FDG = ∠EGD + ∠FGD

Or, ∠EDF = ∠EGF.

That is, ∠BAC = ∠EDF

So in the ∆ABC and ∆DEF, AB = DE, AC = DF and the included

∠BAC = the included ∠EDF.

Therefore, ∆ABC = ∆DEF [by SAS]** (proved).**

### If one angle of a triangle is greater than another, then the side opposite the greater angle is greater than the side opposite the less.

**Particular enunciation: **Let ABC be a triangle in which ∠ABC >∠ACB. It is required to prove that AC > AB.

**Proof: **If AC is not greater than AB, it must be either equal or less than AB i.e., AC=AB or AC<AB.

Now if AC were equal to AB then the ∠ABC would be equal to the ∠ACB.

But by hypothesis, this is not true.

Again, if AC were less than AB then

The ∠ABC would be less than the ∠ACB.

But, by hypothesis this is not true. That is AC is neither equal to, nor less than AB.

∴ AC>AB

#### ABCD is a quadrilateral in which AB = AD, BC = CD, and BC > AB. It is required to proved that ∠DAB> the ∠DCB

**Particular enunciation**: Let ABCD be is a quadrilateral in which AB = AD, BC =AD, BC = CD, and BC > AB. It is required to proved that ∠DAB> the ∠DCB.

Proof: In the ∆ABC, BC > AB,

Therefore, the ∠BAC > the ∠BCA…………….. (i**) **

Again, from the ∆ADC,CD>AD.

Therefore, the ∠DAC > the ∠DCA ………………. (ii)

Adding (1) and (2) we have,

∠BAC> + ∠DAC>∠BCA + ∠DCA

That is, ∠DAB> ∠DCB** (proved).**