Angle Angle Side Postulate

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angle bisector of ∠A and let it meet at D.

Statements	Reasons
1) ∠B = ∠C	1) Given
2) ∠BAD = ∠CAD	2) By construction
3) AD = AD	3) Reflexive (common side)
4) ΔABD ≅ ΔACD	4) By angle angle side postulate (AAS)
5) AB = AC	5) By CPCTC

Examples :

1) If ΔABC is an isosceles triangle with AB = AC. Prove that the perpendiculars from the vertices B and C to their opposite sides are equal.

Given : ΔABC is an isosceles triangle with AB = AC.

Prove that : BD = CE

Statements	Reasons
1) AB = AC	1) Given
2) ∠ABC = ∠ACB	2) If two sides are congruent then the angle opposite to them are also congruent
3) ∠CEB = ∠BDC	3) Each 900
4) BC = BC	4) Reflexive (common side)
5) ΔBCE ≅BCD	5) By AAS postulate
6) BD = CE	6) CPCTC

2) ∠A = ∠C and AB = BC. Prove that ΔABD ≅ ΔCBE.

Given : ∠A = ∠C and AB = BC

Prove that : ΔABD ≅ ΔCBE

Statements	Reasons
1) ∠A = ∠C	1) Given
2) ∠AOE = ∠COD	2) Vertically opposite angles
3) ∠A + ∠AOE = ∠C + ∠COD	3) Add (1) and (2)
4) 1800 - ∠AEO = 1800 - ∠CDO	4) Since ∠A + ∠AOE + ∠AEO = 180 and ∠C + ∠COD + ∠CDO = 180
5) ∠AEO = ∠CDO	5) By subtraction property
6) ∠AEO + ∠OEB = 1800	6) Linear pair angles
7) ∠CDO + ∠ODB = 1800	7) Linear pair angles
8) ∠AEO + ∠OEB = ∠CDO + ∠ODB	8) Transitive property
9) ∠OEB = ∠ODB	9) Subtraction property and from(5)
10) ∠CEB = ∠ADB	10) Since ∠OEB = ∠CEB and ∠ODB = ∠ADB
11) AB = BC	11) Given
11) ΔABD ≅ ΔCBE	11) By AAS postulate (from (1),(10))

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