Given the parallelogram ABCD, draq diagonal AC
From the parallelogram above, it can observed that line BCis parallel to line AD. Also, the line AB is parallel to line DC
The diagonal AC is a transversal to the two parallel lines BC and line AD
In the triangles ABC and triangle CDA,
[tex]\begin{gathered} \angle BAC=\angle DCA(\text{alternate angles of parallel line AB and line DC)} \\ \angle BCA=DAC(alternate\text{ angles of parallel line BC and line AD)} \end{gathered}[/tex]
From the two triangles ABC and CDA, line AC and line CA are common to them
[tex]i\mathrm{}e\text{. AC=CA (Common length)}[/tex]
It can be deduced that triangle ABC is congruent to triangle CDA
[tex]ie\text{. }\Delta ABC\cong\Delta CDA(A\text{ AS)}[/tex]
Hence, AB≅CD and BC≅DA (Matching sides of congruent triangles ABC and triangles CDA)
