**(1) Triangle : It is a closed figure formed by three intersecting lines. It has three sides, three angles and three vertices.
**Consider a triangle ABC shown below:The triangle ABC will be denoted as âˆ† ABC.Â Here, âˆ† ABC have three sides AB, BC, CA; three angles âˆ A, âˆ B, âˆ C and three vertices A, B, C.

**(2) Congruence of Triangles:Â ****The word â€˜ congruentâ€™ means equal in all aspects or the figures whose shapes and sizes are same.**

For triangles, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle then they are said to be congruent triangles.

** For Example:**Â Consider two âˆ† ABC and âˆ† PQR as shown below:Here, âˆ† ABC is congruent to âˆ† PQR which is denoted as âˆ† ABC â‰… âˆ† PQR.

âˆ† ABC â‰… âˆ† PQR means sides AB = PQ, BC = QR, CA = RP; the âˆ A = âˆ P, âˆ B = âˆ Q, âˆ C = âˆ R and vertices A corresponds to P, B corresponds to Q and C corresponds to R.

**(3) Criteria for Congruence of Triangles:
**

*For example*:Â Prove Î” AOD â‰… Î” BOC.From figure, we can see that

OA = OB and OC = OD

Also, we can see that, âˆ AOD and âˆ BOC form a pair of vertically opposite angles,

âˆ AOD = âˆ BOC

Now, since two sides and an included angle of triangle are equal, by SAS congruence rule, we can write that Î” AOD â‰… Î” BOC.

**(ii) ASA Congruence Rule:
**

We need to prove that Î” ABC â‰… Î” DEF.

Case 1: Suppose AB = DE.From the assumption, AB = DE and given that âˆ B = âˆ E, BC = EF, we can say that Î” ABC â‰… Î” DEF as per the SAS rule.

Case 2: Suppose AB > DE or AB < DE.Let us take a point P on AB such that PB = DE as shown in the figure.

Now, from the assumption, PB = DE and given that âˆ B = âˆ E, BC = EF, we can say that Î” PBC â‰… Î” DEF as per the SAS rule.

Now, since triangles are congruent, their corresponding parts will be equal. Hence, âˆ PCB = âˆ DFE

We are given that âˆ ACB = âˆ DFE, which implies that âˆ ACB = âˆ PCB

This thing is possible only if P are A are same points or BA = ED.

Thus, Î” ABC â‰… Î” DEF as per the SAS rule.

On similar arguments, for AB < DE, it can be proved that Î” ABC â‰… Î” DEF.

** For Example:** AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.From the figure, we can see that,

âˆ AOD = âˆ BOC (Vertically opposite angles)

âˆ CBO = âˆ DAO (Both are of 90

BC = AD (Given)

Now, as per AAS Congruence Rule, we can say that Î” AOD â‰… Î” BOC.

Hence, BO = AO which means CD bisects AB.

**(4) Some Properties of a Triangle:
**

We need to prove that âˆ B = âˆ CFirstly, we will draw bisector of âˆ A which intersects BC at point D.

For the Î” BAD and Î” CAD, given that AB = AC, from the figure âˆ BAD = âˆ CAD and AD = AD.

Thus, by SAS rule Î” BAD â‰… Î” CAD.

Therefore, âˆ ABD = âˆ ACD, since they are corresponding angles of congruent triangles.

Hence, âˆ B = âˆ C.

** For Example:**Â In âˆ† ABC, AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.From the figure, we can see that in Î” ABD and Î” ACE,

AB = AC and

âˆ B = âˆ C (Angles opposite to equal sides)

Given that BE = CD.

Subtracting DE from both the sides, we have,

BE â€“ DE = CD â€“ DE i.e. BD = CE.

Now, using SAS rule, we can say that Î” ABD â‰… Î” ACE

Therefore, by CPCT, AD = AE.

** Theorem 2:** The sides opposite to equal angles of a triangle are equal.

It is given that, âˆ BAD = âˆ CAD

AD = AD (Common side)

âˆ ADB = âˆ ADC = 90Â°

So, Î” ABD â‰… Î” ACD by ASA congruence rule.

Therefore, by CPCT, AB = AC (CPCT) or in other words Î” ABC is an isosceles triangle.

**(5) Some More Criteria for Congruence of Triangles:
**

** For Example:Â **Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Î” PQR. Show that Î” ABM â‰… Î” PQN.From the figure, we can see that, AM is the median to BC.

So, BM = Â½ BC.

Similarly, PN is median to QR. So, QN = Â½ QR.

Now, BC = QR.

So, Â½ BC = Â½ QR i.e. BM = QN

Given that, AB = PQ, AM = QN and AM = PN.

Therefore, Î” ABM â‰… Î” PQN by SSS Congruence Rule.

**(6) Inequalities in a Triangle:
**

** For Example:Â **For the given figure, PR > PQ and PS bisects âˆ QPR. Prove that âˆ PSR > âˆ PSQ.Given, PR > PQ.

Therefore, âˆ PQR > âˆ PRQ (As per angle opposite to larger side is larger) â€“ (1)

Also, PS bisects QPR, so, âˆ QPS = âˆ RPS â€“ (2)

Now, âˆ PSR = âˆ PQR + âˆ QPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. â€“ (3)

Similarly, âˆ PSQ = âˆ PRQ + âˆ RPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. â€“ (4)

Adding (1) and (2), we get,

âˆ PQR + âˆ QPS > âˆ PRQ + âˆ RPS

Now, from 3 & 4, we get,

âˆ PSR > âˆ PSQ.

** For Example:Â **D is a point on side BC of Î” ABC such that AD = AC. Show that AB > AD.Given that AD = AC,

Hence, âˆ ADC = âˆ ACD as they are angles opposite to equal sides.

Now, âˆ ADC is an exterior angle for Î”ABD. Therefore, âˆ ADC > âˆ ABD or, âˆ ACD > âˆ ABD or, âˆ ACB > âˆ ABC.

So, AB > AC since side opposite to larger angle in Î” ABC.

In other words, AB > AD (AD = AC).

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