Question

given: ( overline{hi} cong overline{gj} ), ( overline{hi} parallel overline{gj} ) prove: ( \triangle gjh cong \triangle ihj ) sss & sas proofs practice: given: ( overline{ab} cong overline{cd} ), ( overline{ab} parallel overline{cd} ), ( d ) is the midpoint of ( overline{be} ) prove: ( \triangle abd cong \triangle cde ) given: ( overline{km} cong overline{nm} ), ( overline{lk} cong overline{ln} ) prove: ( \triangle lkm cong \triangle lnm )

Explanation:

7.

Step1: Given

Given \( \overline{HI}\cong\overline{GJ}\)

Step2: Given

Given \( \overline{HI}\parallel\overline{GJ}\)

Step3: Alternate - interior angles

If two parallel lines (\(\overline{HI}\) and \(\overline{GJ}\)) are cut by a transversal (\(\overline{HJ}\)), then \(\angle GJH\cong\angle IHJ\) (alternate - interior angles theorem)

Step4: Reflexive property

For any segment \( \overline{HJ}\), \( \overline{HJ}\cong\overline{HJ}\) (reflexive property of congruence)

Step5: SAS (Side - Angle - Side)

In \(\triangle GJH\) and \(\triangle IHJ\), we have \( \overline{GJ}\cong\overline{HI}\) (from step 1), \(\angle GJH\cong\angle IHJ\) (from step 3), and \( \overline{HJ}\cong\overline{HJ}\) (from step 4). So, \(\triangle GJH\cong\triangle IHJ\) by the SAS (Side - Angle - Side) congruence criterion

9.

Step1: Given

Given \( \overline{AB}\cong\overline{CD}\)

Step2: Given

Given \( \overline{AB}\parallel\overline{CD}\)

Step3: Alternate - interior angles

If two parallel lines (\(\overline{AB}\) and \(\overline{CD}\)) are cut by a transversal (\(\overline{BD}\)), then \(\angle ABD\cong\angle CDE\) (alternate - interior angles theorem)

Step4: Definition of mid - point

Since \(D\) is the mid - point of \( \overline{BE}\), then \( \overline{BD}\cong\overline{DE}\) (definition of mid - point: a mid - point divides a segment into two congruent segments)

Step5: SAS (Side - Angle - Side)

In \(\triangle ABD\) and \(\triangle CDE\), we have \( \overline{AB}\cong\overline{CD}\) (from step 1), \(\angle ABD\cong\angle CDE\) (from step 3), and \( \overline{BD}\cong\overline{DE}\) (from step 4). So, \(\triangle ABD\cong\triangle CDE\) by the SAS (Side - Angle - Side) congruence criterion

10.

Step1: Given

Given \( \overline{KM}\cong\overline{NM}\)

Step2: Given

Given \( \overline{LK}\cong\overline{LN}\)

Step3: Reflexive property

For any segment \( \overline{LM}\), \( \overline{LM}\cong\overline{LM}\) (reflexive property of congruence)

Step4: SSS (Side - Side - Side)

In \(\triangle LKM\) and \(\triangle LNM\), we have \( \overline{LK}\cong\overline{LN}\) (from step 2), \( \overline{KM}\cong\overline{NM}\) (from step 1), and \( \overline{LM}\cong\overline{LM}\) (from step 3). So, \(\triangle LKM\cong\triangle LNM\) by the SSS (Side - Side - Side) congruence criterion

Answer:

7.

Statements	Reasons
\(2.\overline{HI}\parallel\overline{GJ}\)	\(2.\) Given
\(3.\angle GJH\cong\angle IHJ\)	\(3.\) Alternate - interior angles theorem
\(4.\overline{HJ}\cong\overline{HJ}\)	\(4.\) Reflexive property
\(5.\triangle GJH\cong\triangle IHJ\)	\(5.\) SAS

9.

Statements	Reasons
\(2.\overline{AB}\parallel\overline{CD}\)	\(2.\) Given
\(3.\angle ABD\cong\angle CDE\)	\(3.\) Alternate - interior angles theorem
\(4.\overline{BD}\cong\overline{DE}\)	\(4.\) Definition of mid - point
\(5.\triangle ABD\cong\triangle CDE\)	\(5.\) SAS

10.

Statements	Reasons
\(2.\overline{LK}\cong\overline{LN}\)	\(2.\) Given
\(3.\overline{LM}\cong\overline{LM}\)	\(3.\) Reflexive property
\(4.\triangle LKM\cong\triangle LNM\)	\(4.\) SSS