In a ΔABC,DE||BC, if AD =x, DB = X- 2, AE = x + 2 and EC - X - 1, then value of x is
In ΔADEandΔABC,∠Aiscommon,∠ADE=∠ABCand∠AED=∠ACB[correspondinganglepairs]
Hence the triangles are similar by AAA.
So, ADAB=AEAC⟹x2x−2=x+22x+1⟹x2x−2=x+22x+1⟹2x2+x=2x2+4x−2x−4⟹x=2x−4orx=4