In the given figure, `Delta CDE ` is an equilateral triangle triangle formed on a side `CD` of a square `ABCD`. Show that `DeltaADE cong Delta BCE.`

Explanation

Given in figure `Delta CDE` is an equilateral triangle formed on a side CD of a square ABCD.
To show `DeltaADE cong Delta BCE `
Proof in `Delta ADE and Delta BCE, " " DE = CE " "["sides of an equliateral triangle"]`
`" "angleADE=angleBCE `
`" "[(because angleADC=angleBCD=90^(@) and angleEDC=angleECD=60^(@)),(therefore angleADE=90^(@)+60^(@)=150^(@) and angleBCE=90^(@)+60^(@)=150^(@))]`
`and " "AD=BC" "["sides of a square"]`
`therefore " " DeltaADE cong Delta BCE " "["by SAS congruence rule "]`