Step: 1

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

Step: 2

As ∠CAB ≅ ∠ACD, AC ¯ = AC ¯ and ∠ACB ≅ ∠CAD, by ASA Postulate, we have ΔACB ≅ ΔCAD.

Correct Answer is : ΔACB ≅ ΔCAD

Step: 1

[Given.]

Step: 2

∠BAC = ∠DEC

[Given.]

Step: 3

∠ACB = ∠ECD

[Vertical angles are congruent.]

Step: 4

Step: 5

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

[AAS postulate.]

Step: 6

Therefore, ΔABC ≅ Δ EDC by AAS postulate.

Correct Answer is : AAS postulate

Step: 1

∠ADB = ∠CDB

[Given.]

Step: 2

∠ABD = ∠CBD

[Given.]

Step: 3

[Reflexive property of congruence.]

Step: 4

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

[ASA postulate.]

Step: 5

Therefore, ΔABD ≅ ΔCBD by ASA postulate.

Correct Answer is : ASA postulate

Step: 1

[PQRS is s square.]

Step: 2

[Given.]

Step: 3

⇒ PQ ¯ = QR ¯ = RS ¯ = SP ¯ = QT ¯

[From Step 1.]

Step: 4

All four angles of a square are equal to 90°. The diagonals of the square bisect its angles.

Step: 5

⇒ ∠OQP = ∠OPQ = ∠OQR = ∠ORQ = ∠ORS = ∠OSR = ∠OSP = ∠OPS = 45°

Step: 6

∠QRT = ∠QTR = 45°

[∠RQT = 90° and QR ¯ = QT ¯ .]

Step: 7

[U is the midpoint of line segment RT.]

Step: 8

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

Step: 9

Therefore, ΔOPQ ≅ ΔOQR ≅ ΔORS ≅ ΔOSP ≅ ΔURQ ≅ ΔUQT.

Correct Answer is : 6

Step: 1

∠PQR = ∠STR

[Given.]

Step: 2

∠PRQ = ∠SRT

[Vertical angles are congruent.]

Step: 3

[Given.]

Step: 4

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

Step: 5

Therefore, ΔPQR ≅ ΔSTR by ASA postulate.

Correct Answer is : ASA postulate

Step: 1

∠CAB = ∠ZXY

[Given.]

Step: 2

∠ABC = ∠XYZ

[Given.]

Step: 3

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

[AAS Theorem.]

Step: 4

The non included sides of ΔABC are AC, BC and that of ΔXYZ are XZ, YZ.

Step: 5

Therefore, ΔABC ≅ ΔXYZ if either AC ¯ = XZ ¯ or BC ¯ = YZ ¯ .

Step: 6

But, only the information BC ¯ = YZ ¯ will be need to prove that ΔABC = ΔXYZ by the AAS theorem.

Correct Answer is : II only

Step: 1

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

[ASA postulate.]

Step: 2

As ∠PSQ ≅ ∠RQS, ∠PQS ≅ ∠RSQ and SQ ¯ = SQ ¯ , by ASA postulate, we have ΔPSQ ≅ΔRQS.

Correct Answer is : ASA postulate

Step: 1

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

Step: 2

As ∠SPQ ≅ ∠QRS, ∠PQS ≅ ∠QSR and SQ ¯ = SQ ¯ , by AAS postulate, we have ΔPQS ≅ ΔRSQ.

Correct Answer is : AAS postulate

Step: 1

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

Step: 2

As ∠ABC ≅ ∠CED, ∠ACB ≅ ∠ECD and AC ¯ = CD ¯ , by AAS theorem, we have ΔABC ≅ ΔDEC

Correct Answer is : AAS postulate

Given :

Step: 1

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

[ASA Theorem.]

Step: 2

As ∠BAC ≅ ∠CED, ∠ABC ≅ ∠CDE and AB ¯ ≅ DE ¯ , by ASA theorem, we have ΔABC ≅ ΔEDC

Correct Answer is : ASA postulate

Step: 1

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

[AAS Theorem.]

Step: 2

As ∠ADC ≅ ∠BDC, ∠CAD ≅ ∠DBC and CD ¯ = CD ¯ , by AAS theorem, we have ΔADC ≅ ΔBDC.

[AAS Theorem.]

Correct Answer is : AAS postulate

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