<Tri-2024-2145>, hint

Let <EAB=<DAB=x, <EAC=<DAC=y.

Since C(1) is tangent to AC at A  <DAC=<DBA=y

C(2) is tangent to AB at A <DAB=<DCA=x. Hence  triangles DAB and DCA are similar

AD/DC=DB/AD---> AD^2=DB x DC

<EDB=<EDC=x+y=<A,  <AEC=ABC, <AEB=<ACB Hence Triangle AEC and ABE are similar

A=DE/DC=DB/DE---> DE^2=DB x DC. Hence DA=DE.