A) \[\Delta BEC\cong \Delta BDC\]
B) \[\Delta AEB\cong \Delta ADC\]
C) \[BC=CD\]
D) None of these
Correct Answer: B
Solution :
We have, AE = AD and GE = BD \[\Rightarrow \]\[AE+CE=AD+BD\] \[\Rightarrow \]\[AC=AB\] ?(i) Now, in \[\Delta AEB\]and \[\Delta ADC,\]we have AE = AD [Given] \[\angle EAB=\angle DAC\] [Common] \[AB=AC\] [From (i)] \[\therefore \]\[\Delta AEB\cong \Delta ADC.\] [By SAS congruency]
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