A) In △ABDsinBAD=sinA3BD⇒BD=sinA3sinB
B) △ADEsinA3DE=sinAEDAD=sin(C+A3)AD
As ∠AED=C+A3 , gives
DE=sinA3sin(C+A3)
C) In △AECsinA3EC=sinAECAC⇒EC=ACsinA3sin(C+A3) ...(1)
And in △ADCsinADCAC=sinCAD⇒AC=sin(B+A3)sinC (2)
From (1) and (2)
EC=sin(B+A3)sinA3sinC
D) In △ADEsinAEDAD=sinADCAE⇒sin(C+A3)AD=sin(B+A3)AE
As ∠AED=C+A3 and ∠ADC=B+A3, gives
AE=sin(B+A3)sin(C+A3)