In ΔABC, ∠C = 54°, the perpendicular bisector of AB at D meets BC at E. If ∠EAC = 42°, then what is the value (in degrees) of ∠ABC?

Given : ED is the perpendicular bisectors of AB, $$\angle C=54^\circ$$ and $$\angle EAC=y=42^\circ$$

To find : $$\angle B=x=?$$

Solution : In $$\triangle$$ EAC, using exterior angle property,

=> $$\angle$$ AEB = $$\angle$$ C + $$y$$

=> $$\angle$$ AEB = $$54^\circ+42^\circ=96^\circ$$

Thus, in $$\triangle$$ AEB, => $$x+z=180^\circ-96^\circ=84^\circ$$ ------------(i)

Also, in $$\triangle$$ EAD and $$\triangle$$ BDE

AD = DB  (DE bisects AB)

$$\angle$$ EAD = $$\angle$$ EDB = $$90^\circ$$

DE = DE   (Common)

Thus, $$\triangle$$ EAD $$\cong$$ $$\triangle$$ BDE    (By SAS criterion)

=> $$x=z$$    (By CPCT)

Substituting above value in equation (i), we get :

=> $$x+x=2x=84^\circ$$

=> $$x=\frac{84}{2}=42^\circ$$

=> Ans - (B)