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practice 1 state the theorem that proves the triangles are congruent. t…

Question

practice
1 state the theorem that proves the triangles are congruent. then, write a congruence statement.
a
b
c
d
©carnegie learn

Explanation:

Part (a)

Step 1: Identify given information

We have two right triangles \( \triangle ABE \) and \( \triangle CBD \). \( \angle A=\angle C = 90^{\circ}\) (right angles), \( AE = CD\) (marked with one tick), and \( AB = CB\) (since \( B \) is on \( AC \), and we can assume \( AB = CB\) or by vertical angles or other, but also \( \angle ABE=\angle CBD\) (vertical angles). Wait, actually, looking at the marks, \( AE = CD\) (one tick), \( \angle A=\angle C = 90^{\circ}\), and \( AB = CB\)? Wait, no, maybe \( EB = DB\)? Wait, no, let's re - examine. The triangles are \( \triangle AEB\) and \( \triangle CDB\). \( \angle A=\angle C = 90^{\circ}\), \( AE = CD\) (marked), and \( AB = CB\)? Wait, no, maybe \( AB = BC\)? Wait, actually, the correct congruence theorem here is AAS or ASA? Wait, no, \( \angle A=\angle C = 90^{\circ}\), \( AE = CD\), and \( \angle ABE=\angle CBD\) (vertical angles). So by AAS (Angle - Angle - Side) or ASA? Wait, \( \angle A=\angle C\), \( \angle ABE=\angle CBD\), and \( AE = CD\), so AAS. Wait, or maybe HL? No, HL is for hypotenuse - leg. Wait, no, \( AE\) and \( CD\) are legs, \( AB\) and \( CB\) are another pair. Wait, maybe ASA. Wait, let's correct. \( \angle A=\angle C = 90^{\circ}\), \( AB = CB\) (if \( B\) is the mid - point), and \( \angle ABE=\angle CBD\) (vertical angles). So by ASA (Angle - Side - Angle), \( \triangle AEB\cong\triangle CDB\)

Step 2: State the theorem and congruence statement

Theorem: ASA (Angle - Side - Angle) or AAS (Angle - Angle - Side). Wait, let's check again. \( \angle A=\angle C = 90^{\circ}\), \( AE = CD\) (side), \( \angle AEB=\angle CDB\) (since the triangles are right - angled and the other angles). Wait, maybe AAS. The congruence statement: \( \triangle AEB\cong\triangle CDB\)

Part (b)

Step 1: Identify given information

We have quadrilateral \( MHTA\) with diagonal \( MT\). \( \angle HMT=\angle ATM\) (marked angles), \( \angle HMT\) and \( \angle ATM\), and \( MT = MT\) (common side), and \( MH = TA\)? Wait, no, looking at the diagram, it's a parallelogram? Wait, \( MH\parallel TA\) and \( MA\parallel HT\). The triangles are \( \triangle MHT\) and \( \triangle TAM\)? No, the triangles are \( \triangle MHT\) and \( \triangle TAM\)? Wait, no, the triangles are \( \triangle MHT\) and \( \triangle TAM\)? Wait, actually, \( \angle HMT=\angle ATM\), \( MT = MT\), and \( \angle HMT\) and \( \angle ATM\), and \( MH = TA\)? Wait, no, the correct congruence theorem here is ASA. \( \angle HMT=\angle ATM\), \( MT = MT\), and \( \angle HMT\) and \( \angle ATM\), and \( MH = TA\)? Wait, no, let's see the angles. \( \angle HMT=\angle ATM\), \( MT\) is common, and \( \angle HMT\) and \( \angle ATM\), and \( MH = TA\)? Wait, maybe SAS. Wait, \( \angle HMT=\angle ATM\), \( MT = MT\), and \( MH = TA\)? No, the diagram shows that \( \angle HMT\) and \( \angle ATM\) are marked, and \( MH\) and \( TA\) are sides. Wait, the correct theorem is ASA. The congruence statement: \( \triangle MHT\cong\triangle TAM\)

Part (c)

Step 1: Identify given information

We have triangles \( \triangle MLN\) and \( \triangle PLN\)? Wait, no, the triangles are \( \triangle MLN\) and \( \triangle PLN\)? Wait, the marks: \( ML = PL\) (one tick), \( MN = PN\) (two ticks), and \( LN = LN\) (common side). So by SSS (Side - Side - Side) congruence theorem.

Step 2: State the theorem and congruence statement

Theorem: SSS (Side - Side - Side). Congruence statement: \( \triangle MLN\cong\triangle PLN\)

Part (d)

Step 1: Identify given information

We have two triangles \( \triangle ABC\) and…

Answer:

Part (a)

Theorem: AAS (or ASA) (Angle - Angle - Side or Angle - Side - Angle). Congruence statement: \( \triangle AEB\cong\triangle CDB\)

Part (b)

Theorem: ASA (Angle - Side - Angle). Congruence statement: \( \triangle MHT\cong\triangle TAM\)

Part (c)

Theorem: SSS (Side - Side - Side). Congruence statement: \( \triangle MLN\cong\triangle PLN\)

Part (d)

Theorem: SAS (Side - Angle - Side). Congruence statement: \( \triangle ABC\cong\triangle EDC\)