**General enunciation: If two triangles have two sides of the one equal to two sides of the other, each to each, the angles included by those sides are also equal then the triangles are equal in all respect.**

**Particular Enunciation:** Let, ∆ABC and ∆DEF be two triangles in which AB=DE, AC=DF and the included ∠BAC= the included ∠EDF.

It is required to prove that ∆ABC≌∆DEF

**Proof: **Apply ∆ABC to the ∆DEF So that the point A falls on the point

D, the side AB along the side DE and c falls on the same side of DE as F

Then because AB=DE

∴the point B must coincide with the point E

And again because AB falls along DE and ∠BAC= ∠EDF

∴AC must fall along DF, Now since AC=DF

∴the point c must coincide with the point F.

Then since B coincides with A and C with F

∴the side BC must coincide with the side EF

Hence the ∆ABC coincides with the ∆DEF.

∴∆ABC≌ ∆DEF (proved)

**General enunciation: **If a side of a triangle is extended, the exterior angle so formed is greater than each of the two interior opposite angles.

**Particular enunciation**: Suppose the side BC of the ∆ABC is extended up to D. It is required to prove that the exterior angle ∠ACD is greater than each of the two interior opposite angles ∠CAB and ∠ABC.

**Construction: **We join B and P, midpoint of AC and extend BP up to E so that PE=BP.

**Proof: **In the triangles ∆PCE and ∆PAB

PC=PA (∵P is the midpoint of AC)

PE=PB (According to construction)

∠CPE= ∠APB (opposite angle)

∆PCE and ∆PAB are congruent.

∴∠PCE=∠PAB

i.e. ∠ACE= ∠CAB …….. (1)

But they are alternative angle of EC and AB with respect to their

intersector.

Hence, EC ǁ AB

So, ∠ECD= ∠ABC ……(2)

But both ∠ACE and ∠ECD are part of ∠ACD

∴From (1) and (2) we get

∠ACD > ∠CAB and ∠ACD> ∠ABC (proved)

**Problem: **In the figure AD is a line segment and B and C lie on the Line AD.

Again, ∠EBC= ∠ECB

BE=CE

AB=CD.

Prove that ∆ABE ≌ ∆DCE.

**Proof**: The line segment BR meets the line AC at the extreme point B.

**∴** ∠RBA+ ∠RBC= 2 right angles ……………… (1)

And the line CE meets the line BD at the extreme point C

∴ ∠RCB +∠RCD= two right angles ……………………(2)

We get from (1) and(2)

∠RBA+∠RBC =∠RCB+∠RCD

Or ∠RBA+∠RBC =∠RBC+∠RCD [since ∠RBC= ∠RCB (given)

Cancelling ∠RBC from both the sides, ∠RBA= ∠RCD.

Now from the ∆ABE and ∆DCR

BR=CR, AB=CD and the included RBA= included RCD

∴ ∆ABR≌ ∆DCR (proved).

**General enunciation**: **If two sides of a triangle are equal, then the angles opposite the equal sides are also equal.**

**Particular enunciation: **Suppose in the ∆ABC, AB = AC. It is required to proved that, ∠ABC =∠ACB

**Construction: **We construct the bisector AD of ∠BAC, which meets BC at D.

**Proof:** In the ∆ABD and ∆ACD.AB = AC , AD is common to both the triangle and the included ∠BAD = the included ∠CAD.

So, ∆ABD ≌∆ ACD

∴ ∠ABD = ∠ACD

The is, ∠ABC = ∠ACB (Proved)

**General enunciation** : If the equal sides of a isosceles triangle are product the exterior or angles so formed are equal.

**Particular enunciation**: Suppose in the ∆ABC, AB = AC. AB and AC are produced to D and E respectively. It is required to proved that ∠DBC = ∠ECB

Proof: The line–Segment BC meets the straight line AD at the point

B.

Therefore, ∠ABC + ∠DBC = 2 right angles

Similarly, ∠ACB + ∠ECB = 2 right angles

So, ∠ABC + ∠DBC = ∠ACB + ∠ECB

Or, ∠ABC + ∠DBC = ∠ABC + ∠ECB,

Cancelling, ∠ABC from both the sides,

∴∠DBC = ∠ECB. **(Proved)**