Side Angle Side Postulate
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Side angle side postulate -> If two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent by side angle side postulate.
AB ≅ DE, BC ≅ EF and ∠B ≅ ∠E
∴ ΔABC ≅ Δ DEF by SAS
Theorem : Angles opposite to two equal sides of a triangle are equal.
Given : Δ ABC in which AB = AC
Prove that : ∠C = ∠B
Construction : Draw the bisector AD of ∠A which meets BC in D.
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| 1) AB = AC | 1) Given |
| 2) AD is a bisector | 2) By construction |
| 3) ∠BAD = ∠CAD | 3) By definition of angle bisector |
| 4) AD = AD | 4) Reflexive (common side) |
| 5) ΔABD ≅ ΔACD | 5) SAS Postulate |
| 6) ∠B = ∠C | 6) CPCTC |
Examples :
1) O is the mid point of AB and CD. Prove that
i) ΔAOC ≅ ΔBOD ii) AC = BD and iii) AC || BD .
Given : O is the mid point of AB and CD.
Prove that : i) ΔAOC ≅ ΔBOD ii) AC = BD and iii) AC || BD .
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| 1) O is the mid point. | 1) Given |
| 2) AO = OB | 2) By definition of mid point. |
| 3) ∠AOC = ∠BOD | 3) Vertically opposite angles |
| 4) CO = OD | 4) By definition of mid point. |
| 5) ΔAOC ≅ ΔBOD | 5) SAS postulate |
| 6) AC = BD | 6) CPCTC |
| 7) ∠CAO = ∠DBO | 7) CPCTC |
| 8) AC || BD | 8) If alternate interior angles are congruent then the lines are parallel. |
2) If D is the mid point of the hypotenuse AC of a right triangle ABC, prove that BD = ½ AC.
Given : ΔABC in which ∠B = 90 0 and D is the mid point of AC.
Prove that : BD = ½ AC.
Construction : Produce BD to E so that BD = DE. Join EC.
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| 1) AD = DC | 1) Given |
| 2) BD = DE | 2) By construction |
| 3) ∠ADB = ∠CDE | 3) Vertically opposite angles |
| 4) ΔADB ≅ ΔCDE | 4) By SAS postulate |
| 5) EC = AB and ∠CED = ∠ABD | 5) CPCTC |
| 6) CE || AB | 6) If alternate interior angles are congruent then the lines are parallel |
| 7) ∠ABC + ∠ECB = 180 | 7) Angles formed on the same side of transveral are supplementary. |
| 8) 90 + ∠ECB = 180 | 8) Since ∠B = 90 given |
| 9) ∠ECB = 180 -90 = 90 | 9) By subtraction property |
| 10) AB = EC | 10) From (5) |
| 11) BC = CB | 11) Reflexive (Common side) |
| 12) ∠ABC = ∠ECB | 12) Each 900 |
| 13) ΔABC ≅ ΔECB | 13) SAS postulate |
| 14) AC = BE | 14) CPCTC |
| 15) 1/2AC = 1/2BE ⇒ 1/2AC = BD | 15) Multiply by 1/2 but 1/2BE = BD by mid point definition. |
• Side Angle Side Postulate
• Side Side Side Postulate
• Angle Angle Side Postulate
• Angle Side Angle Postulate
• HL postulate (Hypotenuse – Leg OR RHS)
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