Triangles
Triangle
A closed figure with three sides is called a Triangle. It has three vertex, sides and Angles.
Types of Triangle
1. There are three types of triangles on the basis of the length of the sides.
| Name of Triangle | Property | Image |
| Scalene | Length of all sides are different |  |
| Isosceles | Length of two sides are equal |  |
| Equilateral | Length of all three sides are equal |  |
2. There are three types of triangles on the basis of angles.
| Name of Triangle | Property | Image |
| Acute | All the three angles are less than 90° |  |
| Obtuse | One angle is greater than 90° |  |
| Right | One angle is equal to 90° |  |
Congruence
If the shape and size of two figures are same then these are called Congruent.
1. Two circles are congruent if their radii are same.
2. Two squares are congruent if their sides are equal.
Congruence of Triangles
A triangle will be congruent if its corresponding sides and angles are equal.
The symbol of congruent is “≅”.
AB = DE, BC = EF, AC = DF
m∠A = m∠D, m∠B = m∠E, m∠C = m∠F
Here ∆ABC ≅ ∆DEF
Criteria for Congruence of Triangles
| S.No. | Rule | Meaning | Figure |
| 1. |
SAS (Side-Angle-Side) Congruence rule |
If the two sides and the including angle of one triangle is equal to another triangle then they are called congruent triangles. |
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| 2. |
ASA (Angle-Side-Angle) Congruence rule |
If the two angles and the including side of one triangle is equal to another triangle then they are called congruent triangles. |
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| 3. |
AAS (Angle-Angle-Side) Congruence rule |
If any two pairs of angles and a pair of the corresponding side is equal in two triangles then these are called congruent triangles. |
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| 4. |
SSS (Side-Side-Side) Congruence rule |
If all the three sides of a triangle are equal with the three corresponding sides of another triangle then these are called congruent triangles. |
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| 5. |
RHS (Right angle-Hypotenuse-Side) Congruence rule |
If there are two right-angled triangles then they will be congruent if their hypotenuse and any one side are equal. |
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Remark
1. SSA and ASS do not show the congruency of triangles.
2. AAA is also not the right condition to prove that the triangles are congruent.
Criteria for Congruence of Triangles:
(i) SAS Congruence Rule:
Statement: Two triangles are congruent if two sides and the included angle of one triangle are equal to the sides and the included angle of the other triangle.
For example: Prove Δ AOD ≅ Δ BOC.
From figure, we can see that
OA = OB and OC = OD
Also, we can see that, ∠ AOD and ∠ BOC form a pair of vertically opposite angles,
∠ AOD = ∠ BOC
Now, since two sides and an included angle of triangle are equal, by SAS congruence rule, we can write that Δ AOD ≅ Δ BOC.
(ii) ASA Congruence Rule:
Statement: Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
Proof: Suppose we have two triangles ABC and DEF, such that ∠ B = ∠ E, ∠ C = ∠ F, and BC = EF.
We need to prove that Δ ABC ≅ Δ DEF.
Case 1: Suppose AB = DE.
From the assumption, AB = DE and given that ∠ B = ∠ E, BC = EF, we can say that Δ ABC ≅ Δ 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 ∠ B = ∠ E, BC = EF, we can say that Δ PBC ≅ Δ DEF as per the SAS rule.
Now, since triangles are congruent, their corresponding parts will be equal. Hence, ∠ PCB = ∠DFE
We are given that ∠ ACB = ∠ DFE, which implies that ∠ ACB = ∠PCB
This thing is possible only if P are A are same points or BA = ED.
Thus, Δ ABC ≅ Δ DEF as per the SAS rule.
On similar arguments, for AB < DE, it can be proved that Δ ABC ≅ Δ 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,
∠ AOD = ∠ BOC (Vertically opposite angles)
∠ CBO = ∠ DAO (Both are of 90o)
BC = AD (Given)
Now, as per AAS Congruence Rule, we can say that Δ AOD ≅ Δ BOC.
Hence, BO = AO which means CD bisects AB.
(iii) SSS Congruence Rule:
Statement: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.
For Example: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that Δ ABM ≅ Δ PQN.
From the figure, we can see that, AM is the median to BC.
So, BM = ½ BC.
Similarly, PN is median to QR. So, QN = ½ QR.
Now, BC = QR.
So, ½ BC = ½ QR i.e. BM = QN
Given that, AB = PQ, AM = QN and AM = PN.
Therefore, Δ ABM ≅ Δ PQN by SSS Congruence Rule.
Example
Find the ∠P, ∠R, ∠N and ∠M if ∆LMN ≅ ∆PQR.
Solution
If ∆ LMN ≅ ∆PQR, then
∠L=∠P
∠M =∠Q
∠N =∠R
So,
∠L=∠P = 105°
∠M =∠Q = 45°
∠M + ∠N + ∠L = 180° (Sum of three angles of a triangle is 180°)
45° + 105° + ∠N = 180°
∠N = 180°- 45° + 105°
∠N = 30°
∠N = ∠R = 30°
Some Properties of a Triangle
If a triangle has two equal sides then it is called an Isosceles Triangle.
1. Two angles opposite to the two equal sides of an isosceles triangle are also equal.
2. Two sides opposite to the equal angles of the isosceles triangle are also equal. This is the converse of the above theorem.
Some Properties of a Triangle:
Theorem 1: Angles opposite to equal sides of an isosceles triangle are equal.
Proof: Suppose we are given isosceles triangle ABC having AB = AC.
We need to prove that ∠ B = ∠C
Firstly, we will draw bisector of ∠ A which intersects BC at point D.
For the Δ BAD and Δ CAD, given that AB = AC, from the figure ∠ BAD = ∠ CAD and AD = AD.
Thus, by SAS rule Δ BAD ≅ Δ CAD.
Therefore, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles.
Hence, ∠ B = ∠C.
For Example: In ∆ 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 Δ ABD and Δ ACE,
AB = AC and
∠ B = ∠ C (Angles opposite to equal sides)
Given that BE = CD.
Subtracting DE from both the sides, we have,
BE – DE = CD – DE i.e. BD = CE.
Now, using SAS rule, we can say that Δ ABD ≅ Δ ACE
Therefore, by CPCT, AD = AE.
Theorem 2: The sides opposite to equal angles of a triangle are equal.
For Example: In Δ ABC, the bisector AD of ∠ A is perpendicular to side BC. Show that AB = AC and Δ ABC is isosceles.
From the figure, we can see that in Δ ABD and Δ ACD,
It is given that, ∠ BAD = ∠ CAD
AD = AD (Common side)
∠ ADB = ∠ ADC = 90°
So, Δ ABD ≅ Δ ACD by ASA congruence rule.
Therefore, by CPCT, AB = AC (CPCT) or in other words Δ ABC is an isosceles triangle.
Inequalities in a Triangle
Theorem 1: In a given triangle if two sides are unequal then the angle opposite to the longer side will be larger.
a > b, if and only if ∠A > ∠B
Longer sides correspond to larger angles.
Theorem 2: In the given triangle, the side opposite to the larger angle will always be longer. This is the converse of above theorem.
Theorem 3: The sum of any two sides of a triangle will always be greater than the third side.
Example
Show whether the inequality theorem is applicable to this triangle or not?
Solution
The three sides are given as 7, 8 and 9.
According to inequality theorem, the sum of any two sides of a triangle will always be greater than the third side.
Let’s check it
7 + 8 > 9
8 + 9 > 7
9 + 7 > 8
This shows that this theorem is applicable to all the triangles irrespective of the type of triangle.
For Example: For the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.
Given, PR > PQ.
Therefore, ∠ PQR > ∠ PRQ (As per angle opposite to larger side is larger) – (1)
Also, PS bisects QPR, so, ∠ QPS = ∠ RPS – (2)
Now, ∠ PSR = ∠ PQR + ∠ QPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. – (3)
Similarly, ∠ PSQ = ∠ PRQ + ∠ RPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. – (4)
Adding (1) and (2), we get,
∠ PQR + ∠ QPS > ∠ PRQ + ∠ RPS
Now, from 3 & 4, we get,
∠ PSR > ∠ PSQ.
For Example: D is a point on side BC of Δ ABC such that AD = AC. Show that AB > AD.
Hence, ∠ ADC = ∠ ACD as they are angles opposite to equal sides.
Now, ∠ ADC is an exterior angle for ΔABD. Therefore, ∠ ADC > ∠ ABD or, ∠ ACD > ∠ ABD or, ∠ ACB > ∠ ABC.
So, AB > AC since side opposite to larger angle in Δ ABC.
In other words, AB > AD (AD = AC).
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