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If two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.

If Angle ∠A≅∠D,

Side AC-≅DF, and

Angle ∠C≅∠F,

then ∆ABC≅∆DEF

If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of a second triangle, then the two triangles are congruent.

Given: ∠B=∠E, ∠C=∠F and AC=DF

To prove: ΔABC ≅ ΔDEF

Proof:

In ∆ABC

∠A+∠B+∠C=180° _____________ (1) (By angle sum property)

In ∆DEF ∠D+∠E+∠F=180°_____________ (2) (By angle sum property)

From (1) and (2),

∠A+∠B+∠C= ∠D+∠E+∠F

∠A+∠E+∠F=∠D+∠E+∠F (Given ∠B=∠E and ∠C=∠F)

⇒∠A=∠D _____________________ (3)

Now, in ΔABC and ΔDEF

∠A=∠D (from (3))

AC = DF (Given)

∠C=∠F (Given)

∴ΔABC≅ΔDEF (AAS congruency)

Hence proved.

A flow proof is a step of proof to be written for a theorem. A flow proof uses arrows to show the flow of a logical argument.

Example 1: Prove the Angle-Angle-Side congruence theorem for the given figures.

Solution:

Given: ∠A≅∠D, ∠C≅∠F and BC≅EF

To prove: ΔABC≅ΔDEF

Example 2: In the diagram,CE ⊥ BD and ∠CAB≅∠CAD.

Write a flow proof to show that ΔABE≅ΔADE.

Solution:

Given: CE−⊥BD− and ∠CAB≅∠CAD

To prove: ΔABE≅ΔADE

Proof:

Example 3: OB is the bisector of ∠AOC, PM

⊥ OA and PN ⊥ OC. Show that ∆MPO ≅ ∆NPO.

Solution:

In ∆MPO and ∆NPO

PM ⊥ OM and PN ⊥ ON

⇒∠PMO = ∠PNO = 90°

Also, OB is the bisector of ∠AOC

Then ∠MOP = ∠NOP

OP = OP common

∴ ∆MPO ≅ ∆NPO (By AAS congruence postulate)

Example 4: Prove that △CBD ≅ △ABD from the given figure.

Solution:

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

To prove: △CBD ≅ △ABD

Proof:

BD- is an angle bisector of ∠CDA, ∠C ≅ ∠A (Given)

∠CDB ≅ ∠ADB (By angle bisector)

DB ≅ DB (Reflexive property)

△CBD ≅ △ABD (Definition of AAS)

Hence proved.

Example 5: Prove that △ABD ≅ △EBC in the given figure.

Solution:

Given: AD∥EC, BD≅BC

To prove: △ABD ≅ △EBC

Proof:

We have learned five methods for proving that the triangles are congruent.

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