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Given: parallelogram EFGH

Prove: EG¯¯¯¯¯ bisects HF¯¯¯¯¯¯ and HF¯¯¯¯¯¯ bisects EG¯¯¯¯¯ .

Johnnyboy  Dec 11, 2017

Best Answer 

 #1
avatar+1965 
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parallelogram EFGH

Given

\(\overline{EF}\cong\overline{HG}\)Property of a Parallelogram (If a quadrilateral is a parallelogram, then its opposite sides are congruent)
\(\overline{EF}\parallel\overline{HG}\)Definition of a Parallelogram (A quadrilateral that has opposite sides parallel is a parallelogram)
\(\textcolor{red}{\angle FEG\cong\angle HGE\\ \angle EFH\cong\angle FHG}\)Alternate Interior Angles Theorem
\(\triangle EKF\cong\triangle GKH\)ASA Congruence Postulate
\(\textcolor{red}{\overline{EK}\cong\overline{KG}\\ \overline{FK}\cong\overline{KH}}\)CPCTC
  
TheXSquaredFactor  Dec 11, 2017
edited by TheXSquaredFactor  Dec 11, 2017
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 #1
avatar+1965 
+1
Best Answer
parallelogram EFGH

Given

\(\overline{EF}\cong\overline{HG}\)Property of a Parallelogram (If a quadrilateral is a parallelogram, then its opposite sides are congruent)
\(\overline{EF}\parallel\overline{HG}\)Definition of a Parallelogram (A quadrilateral that has opposite sides parallel is a parallelogram)
\(\textcolor{red}{\angle FEG\cong\angle HGE\\ \angle EFH\cong\angle FHG}\)Alternate Interior Angles Theorem
\(\triangle EKF\cong\triangle GKH\)ASA Congruence Postulate
\(\textcolor{red}{\overline{EK}\cong\overline{KG}\\ \overline{FK}\cong\overline{KH}}\)CPCTC
  
TheXSquaredFactor  Dec 11, 2017
edited by TheXSquaredFactor  Dec 11, 2017

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