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Home / Geometry / 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.

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.

 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).

 

 

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