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Triangles by Sides and by Angles

Sep 14, 2022

Key Concepts

• Use congruent triangles

Introduction:

Use congruent triangles

Example 1: In the given figure, AB = BC and AD = CD. Show that BD bisects AC at right angles.

Solution:

From the given figure, ∆ABD ≅ ∆CBD

Given: AB = BC and AD = CD

To prove: ∠BEA = ∠BEC = 90° and AE = EC.

Proof:

AB = BC                                                (Given)

BD = BD                                                (Common sides)

Therefore, ∆ABD ≅ ∆CBD                      (By SSS congruency)

∠ABD = ∠CBD                                      (Corresponding angles)

Now, from ∆ABE and ∆CBE,

AB = BC                                                (Given)

∠ABD = ∠CBD                                      (Corresponding angles)

BE = BE                                                (Common sides)

Therefore, ∆ABE≅ ∆CBE                       (By SAS congruency)

∠BEA = ∠BEC                                      (Corresponding angles)

And ∠BEA +∠BEC = 180°                      (Linear pair)

2∠BEA = 180°                                       (∠BEA = ∠BEC)

∠BEA =

180°2180°2

= 90° = ∠BEC

AE = EC                                                (Corresponding sides)

Hence, BD

⊥⊥

AC.

Example 2: In a Δ ABC, if AB = AC and ∠ B = 70°, find ∠ A.

Solution:

Given: AB = AC and ∠B = 70°

∠ B = ∠ C [Angles opposite to equal sides are equal]

Therefore, ∠ B = ∠ C = 70°

Sum of angles in a triangle = 180°

∠ A + ∠ B + ∠ C = 180°

∠ A + 70° + 70° = 180°

∠ A = 180° – 140°

∠ A = 40°.

Example 3: In the given figure, PQ = PS and QPR = SPR. Prove that PQR PSR, and use the SAS congruence postulate.

Solution:

In QPR and PSR,

PQ = PS                        (Given)

∠QPR = ∠SPR              (Given)

PR = PR                        (Common sides)

Therefore, ∆PQR ≅ ∆PSR  (By SAS congruence).

Example 4: Identify the congruent triangle in the given figure.

Solution:

In ∆LMN,

65° + 45° + ∠L = 180°

110° + ∠L = 180°

∠L = 180° – 110°

Therefore, ∠L = 70°

Now in ∆XYZ and ∆LMN

∠X = ∠L       (Given)

XY = LM      (Given)

XZ = NL      (Given)

Therefore, ∆XYZ ≅ ∆LMN by SAS congruence postulate.

Example 5:

Write a 2-column proof for the given figure.

Given: BD is an angle bisector of CDA, C A

To prove: △CBD ≅ ∠ABD

Solution:

How to prove construction:

The following steps explain the construction of congruent triangles:

Step 1:

To copy A, draw a segment starting at point D. Draw an arc with the center A. Using the same radius, draw an arc with center D. Label points B, C, and E.

Step 2:

Draw an arc with radius BC and center E. Label the intersection F.

Step 3:

Draw

DF−→−DF→

.

Example 6:

Write a proof to verify that the construction for copying an angle is valid.

Solution:

BC−BC-

and

EF−EF- to the diagram. In the construction,

AB−AB-,

DE−DE-,

AC−AC-, and

DF−DF- are determined by the same compass. So, the required construction is

BC−BC- and

EF−EF-.

Given:

AB−AB-

DE−DE-,

AC−AC- ≅

DF−DF-,

BC−BC- ≅

EF−EF-.

To prove:

∠∠

D ≅

∠∠A

Plan for Proof:

Show that △CAB ≅ △FDE, so we can conclude that the corresponding parts ∠A and ∠D are congruent.

Plan in action:

Exercise

1. Prove that FL O HN in the given diagram.

2. Prove that APUX – AQSY in the given figure.

3. Prove that AC = GE in the given diagram.

4. Write a two-column proof from the given diagram.

5. Prove that 21 22 from the given diagram with the given information. Given: MNKN, ZPMN

6. Prove that Z1 Z2 from the given diagram with the given information. Given: TS TV, SR_VW 1 RA

7. Find the measure of each angle in the given triangle. m2A=xo;m_B=(4x)”and m_C=(5x)”.

8 Find the measure of each angle in the given triangle. m2A=xo;mB=(5x)’and m2C=(x+19)o.

What have we learned:

• Understand and apply the SSS congruence postulate.
• Understand and apply SAS congruence postulate.
• Understand and apply the AAS congruence postulate.
• Understand and apply construction proof.
• Solve problems on different congruence of triangles.
• Solve problems on different congruence postulates.

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