Given, AE = AD

and CE = BD

On adding Eqs. (i) and (ii), we get

AE + CE = AD + BD

=> AC = AB

Now, in ∆AEB and ∆ADC, we have

AE = AD [given]

AB = AC [proved above]

and ∠A = ∠A [common angle]

∴ ∆AEB =∆ADC [by SAS congruence rule]

Given, AE = AD

and CE = BD

On adding Eqs. (i) and (ii), we get

AE + CE = AD + BD

=> AC = AB

Now, in ∆AEB and ∆ADC, we have

AE = AD [given]

AB = AC [proved above]

and ∠A = ∠A [common angle]

∴ ∆AEB =∆ADC [by SAS congruence rule]