Given, a parallelogram ABCD in which AB=10 cm and AD = 6 cm.
Now, draw a bisector of `angleA` meets DC in E ang produce it to F and produce BC to meet at F.
Also, produce AD to H and join HF, so that ABFH, is a parallelogram.
Since, `" "HF||AB" "`
`therefore" "angleAFH=angle FAB" "`[alternate interior angles]
`" "angleHAF=angleFAB" " ` [since, AF is the bisector of `angle`A]
`rArr" "angleHAF=angleAFH" "` [ from Eq. (i)]
`rArr" "HF=AH" "`[sides opposite to equal angles are equal]
But `" "HF=AB= 10` cm
`therefore" "AH=HF= 10` cm
`rArr" "AD+DH= 10` cm
`rArr" "DH=(10-6)` cm
`therefore" "DH=4` cm
Since, CFHD is a parallelogram.
Therefore, opposite sides are equal.
`therefore" "DH=CF=4` cm
