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In figure ,D and E are Points on side BC...

In figure ,`D` and `E` are Points on side `BC` of a `Delta ABC` such that `BD`=`CE` and `AD`=`AE`. Show that `Delta ABD cong Delta ACE.`

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Given D and E are the point on side BC of a `Delta ABC `Such that BD =CE and AD=AE.
To show `Delta ABD cong Delta ACE`
Proof We have,`" " AD=AE " "[given]`
`rArr " "angleADE=angleAED " "…(i)`
`" "["since angle opposite to equal sides are equal "]`
We have `" "angleADB+angleADE=180 ^(@) " " ["linear Pair axiom"] `
`rArr " "angleADB=180^(@)-angleAED`
`" "=180^(@)-angleAED " " ["from Eq.(i)"]`
`In Delta ABD and Delta ACE,`
`" "angleADB=angleAEC " "[because angleAEC + angleAED=180^(@),"Linear Pair axiom"]`
`" "BD=CE " "["given"]`
`therefore " "DeltaABD cong Delta ACE " "["by SAS congruence rule"]`
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