**General enunciation: If three sides of a triangle are respectively equal to the corresponding three sides of another triangle, then the triangle are congruent.**

**Particular enunciation**: Let, in ∆ABC and ∆DEF, AB = PQ, AC = PR and BC = QR. It is required to prove that, ∆ABC ≌ ∆PQR.

**Proof:** Let BC and PQ are the greatest sides of ∆ABC and ∆PQR respectively.

Let us place ∆ABC upon ∆PQR in such a way that B falls upon Q, the side BC lies along the side QR and the point A falls on the opposite of the side at which side of PQ, S lies. Now, since BC = PQ, The point C falls upon the point R.

Let the new position of the point A be denoted by S. As result ∆PQS is the new position of ∆ABC.

Hence**,** QS= BC, PS = CA and ∠QSP = ∠ACB.

Let us join D, G.

Now in ∆QRS, QR= QS **[∵**QS= BC** = **QR]

∴∠QRS = ∠QSR.

Again, in ∆PRS, PS =PR, [∵PS = CA = PR]

∴∠PRS = ∠PSR.

Hence, ∠QRS + ∠PRS = ∠QSR + ∠PSR

or ∠PRQ = ∠PSQ

i.e. ∠ACB = ∠PRQ.

Therefore, in ∆ABC and ∆PQR, AB = PQ, AC = QR

And included ∠ACB = included ∠PRQ

∴∆ABC ≌∆PQR. (Proved)

**General enunciation: If one side is greater than another side of a triangle, the angle opposite to the greater side is greater than the angle opposite to the smaller side. **

**Particular enunciation:** Let**, **in ∆ABC, AC>AB

It is required to proved that, ∠ABC> ∠BCA.

**Construction:** Let us cut off AD from AC

Such that, AD = AB and join B and D.

**Proof:** In ∆ABD, DA = AB

∴∠ABD = ∠ADB.

But in ∆BDC, exterior ∠ADB> ∠BCD

But ∠ABC>∠ABD. [∵∠ABD is a part of ∠ABC.]

Hence, ∠ABC > ∠ACB (Proved)

**General enunciation: If two angles and a side of a triangle are equal to the corresponding angles and side of another triangle respectively, then two triangles are congruent.**

**Particular enunciation:** Let, in ∆ABC and ∆DEF, ∠A=∠D, ∠B=∠E** **

And the side BC= corresponding side EF.

It is required to prove that, ∆ABC ≌∆DEF.

**Proof**: We know, the sum of three angles is equal to two right angles.

So, in ∆ABC, ∠A+∠B+∠C=180°.

And in ∆DEF, ∠D+∠E+∠F=180°.

∴∠A+∠B+∠C=∠D+∠E+∠F.

∵∠A=∠D and ∠B= ∠E,

∴∠C=∠F.

Now, place ∆ABC on ∆DEF in such a way that the point B lies on E ,the side BC lies alone EF and the point A falls on that side of EF , D lies .Since, BC =EF, therefore, the point C falls on the point F. Now since ∠B=∠E, the side BA will lie alone the side ED and since ∠C=∠F , the side CA will lie alone FD .Hence , the common point A of the side BA and CA will fall on the common point D of the side ED and FD.

Hence, ∆ABC is superimposed on ∆DEF.

∴∆ABC≌∆DEF (proved)

General enunciation: If two hypotenuses of two right angled triangles are equal to each other and also side of one of those triangles is equal to the corresponding side of another, then two triangles are congruent.

**Particular enunciation**: Let, in the right angle triangles ABC and DEF, the hypotenuse BC= hypotenuse DF and AB= OD. It is required to prove that, ∆ABC≌∆DOF.

**Proof**: Let us place ∆ABC upon ∆DOF in such way that the point A lies on the point O. The side BA lies alone the side OD and the point C falls on the opposite of that side at which side of DF, E lies. Let, the new position of the point C is denoted by E.

Hence, ∆DOE is the new position of ∆ABC.

Therefore, DE = BC = DF,

∠DEO=∠ACB and ∠DEO = ∠BAC = one right angle.

Again, since ∠DOF + ∠ DOE= one right angle + one right angle

= Two right angle,

∴ OF and OF lie on the same straight

Now, since in ∆DEF, DE = DF, therefore ∠DFO = ∠DEO.

Hence, ∠DFO = ∠ACB.

Now in ∆ABC and ∆DEO , ∠ABC=∠DOF, [each being one right angle]

∠ACB=∠DFO and side AB = corresponding side DO.

∴∆ABC≌∆DOF (proved)