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Congruent Triangles
• Congruent triangles have
three congruent sides and
three congruent angles.
• However, triangles can be
proved congruent without
showing 3 pairs of
congruent sides and angles.
LAHALLHL
FOR RIGHT TRIANGLES ONLY
AASASASASSSS
FOR ALL TRIANGLES
30°
30°
Q1. Why aren’t these triangles congruent?
What do we call these triangles?
So, how do we prove that triangles are
congruent?
1. A   D
2. AB  DE
3.  B   E
ABC   DEF
C
1. A  D
2. B  E
3. BC  EF
ABC   DEF
1. AB  DE
2. BC  EF
3. AC  DF
ABC   DEF
1. AB  DE
2. A  D
3. AC  DF
ABC   DEF
F
If two isosceles triangles have a common line segments joining their
vertices bisects the common base at right angles.
Given Two
ABC and DBC with the same base BC, in which AB=AC and
DB=DC. Also ,AD meets BC in E.
To Prove- BE=CE and AEB= AEC=90*
A
B
C
D
E
1 2
3 4
B
A
CE
3 4
1 2
In ABD & ACD, we have :
AB=AC (given)
DB=DC(given)
AD=AD(common)
ABD= ACD(SSS criteria)
Now, in ABE & ACE, we have:
AB=AC (given) AE=AE(common)
ABE= ACE (SAS criteria ) BE=CE (.) & 3=4
But 3= 4=180*(linear pair)
24=180*,i.e.4=90*
3=4=90*
Hence, BE=CE & AEB =AEC=90*
B
D
A
CE
3 4
1 2
A
B C
D
E
1 2
3 4
THUS PROVED
Geometry the congruence of triangles

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Geometry the congruence of triangles

  • 1.
  • 2. Congruent Triangles • Congruent triangles have three congruent sides and three congruent angles. • However, triangles can be proved congruent without showing 3 pairs of congruent sides and angles.
  • 3. LAHALLHL FOR RIGHT TRIANGLES ONLY AASASASASSSS FOR ALL TRIANGLES
  • 4. 30° 30° Q1. Why aren’t these triangles congruent? What do we call these triangles? So, how do we prove that triangles are congruent?
  • 5. 1. A   D 2. AB  DE 3.  B   E ABC   DEF C
  • 6. 1. A  D 2. B  E 3. BC  EF ABC   DEF
  • 7. 1. AB  DE 2. BC  EF 3. AC  DF ABC   DEF
  • 8. 1. AB  DE 2. A  D 3. AC  DF ABC   DEF F
  • 9. If two isosceles triangles have a common line segments joining their vertices bisects the common base at right angles. Given Two ABC and DBC with the same base BC, in which AB=AC and DB=DC. Also ,AD meets BC in E. To Prove- BE=CE and AEB= AEC=90* A B C D E 1 2 3 4 B A CE 3 4 1 2
  • 10. In ABD & ACD, we have : AB=AC (given) DB=DC(given) AD=AD(common) ABD= ACD(SSS criteria) Now, in ABE & ACE, we have: AB=AC (given) AE=AE(common) ABE= ACE (SAS criteria ) BE=CE (.) & 3=4 But 3= 4=180*(linear pair) 24=180*,i.e.4=90* 3=4=90* Hence, BE=CE & AEB =AEC=90* B D A CE 3 4 1 2 A B C D E 1 2 3 4 THUS PROVED