It is given that
△ABE≅△ACD
∴ AB=AC [∴ Corresponding parts of congruent triangles are equal]
and, AE=AD
⇒ ABAD=ACAE
⇒ ABAC=ADAE.......(i)
Thus, in triangles ADE and ABC, we have
ABAC=ADAE [from (i)]
and, ∠BAC=∠DAE [Common]
Hence, by SAS-criterion of similarity, we have
△ADE∼△ABC [Hence proved]
![1031824_1009322_ans_76546f5a93c8483e979d55bc74a0c14e.png](https://search-static.byjusweb.com/question-images/toppr_ext/questions/1031824_1009322_ans_76546f5a93c8483e979d55bc74a0c14e.png)