Given that ABCD is a parallelogram, prove that
[tex]\begin{gathered} \bar{AB}\cong\bar{CD} \\ \bar{BC}=\bar{DA} \end{gathered}[/tex]
step 1: Sketch the parallelogram
step 2: The diagonal AC, divides the parallelogram into two triangles
[tex]\begin{gathered} \Delta ADC\text{ and }\Delta ABC \\ Note\text{ that,} \\ \angle DAC=\angle ACB\text{ ( the angles are alternate)} \\ \angle DCA=\angle BAC\text{ (the angles are alternate)} \\ side\text{ AC = AC (common sides for both triangles)} \end{gathered}[/tex]
step 3: By the ASA (Angle-Side-Angle) congruency theorem,
[tex]\begin{gathered} \Delta ADC\cong\Delta ABC \\ (The\text{ two triangles are congruent)} \end{gathered}[/tex]
Hence, by CPCT (corresponding parts of congruent triangles)
[tex]\begin{gathered} \bar{AB}\cong\bar{CD}\text{ } \\ \bar{BC}\cong\bar{DA} \end{gathered}[/tex]
