Question

problem 23
given: ( \triangle bcdcong\triangle efd ),
b is the midpoint
of ( overline{ac} ).
prove: abef is a
parallelogram.

1. ( \triangle bcdcong\triangle efd )
2. click here to insert
3. b is the midpoint of ( overline{ac} ).
4. click here to insert
5. given
6. corresponding parts

of congruent triangles
are congruent
(c.p.c.t.c.)

1. click here to insert
2. click here to insert

Explanation:

Step1: Use CPCTC

Since \(\triangle BCD\cong\triangle EFD\), by Corresponding Parts of Congruent Triangles are Congruent (CPCTC), \(BC = EF\).

Step2: Use mid - point property

Given \(B\) is the mid - point of \(\overline{AC}\), so \(AB=BC\).

Step3: Substitute

From steps 1 and 2, \(AB = EF\) (by substitution).

Step4: Use another property of congruent triangles

Also, from \(\triangle BCD\cong\triangle EFD\), \(\angle CBD=\angle EFD\). So \(AB\parallel EF\) (alternate interior angles are equal implies parallel lines).

Answer:

Since \(AB = EF\) and \(AB\parallel EF\), by the definition of a parallelogram (a quadrilateral with one pair of opposite sides both equal and parallel), \(ABEF\) is a parallelogram.