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.
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.
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).