Question

Solving a proof I know how to start it just confused how to put it all together

Answer 1

Given that ABCD is a parallelogram, prove that

`\begin{gathered} \bar{AB}\cong\bar{CD} \\ \bar{BC}=\bar{DA} \end{gathered}`

step 1: Sketch the parallelogram

step 2: The diagonal AC, divides the parallelogram into two triangles

`\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}`

step 3: By the ASA (Angle-Side-Angle) congruency theorem,

`\begin{gathered} \Delta ADC\cong\Delta ABC \\ (The\text{ two triangles are congruent)} \end{gathered}`

Hence, by CPCT (corresponding parts of congruent triangles)

`\begin{gathered} \bar{AB}\cong\bar{CD}\text{ } \\ \bar{BC}\cong\bar{DA} \end{gathered}`