If the areas of two similar triangles are equal, prove that they are congruent.

Given:Ar(ΔABC)=Ar(ΔDEF)Also,ΔABC∼ΔDEFTo Prove:ΔABC≅ΔDEFConstruction:Draw AO Perpendicular to BC and DP Perpendicular to EFProof:Since,ΔABC∼ΔDEF∴∠A=∠D,∠B=∠E,∠C=∠F(Corresponding Angles of Similar Triangles)(1)Also,AB/DE=BC/EF=AC/DF(Corresponding Sides of Similar Triangles)In ΔAOB and ΔDPE,∠AOB=∠DPB=90o∠B=∠E(From 1)∴By AA Criterion of Similarity,ΔAOB∼ΔDPE∴AB/DE=AO/DP(2)Ar(ΔABC)/Ar(ΔDEF)=(1/2×BC×AO)/(1/2×EF×DP)=1=BC×AO/EF×DP=1∴BC×AO=EF×DPBC/EF=DP/AOBC/EF=AB/DE=AC/DF=AO/DP(using(2))Since,BC/EF=1 and BC×AO=EF×DP, we have AO=DPTherefore, BC=EF, AB=DE, AC=DF, AO=DP using(2)∴ΔABC≅ΔDEF