If two angles of triangles are equal, then the sides opposite to the equal angles are equal.
Particular enunciation: Let ABC be a triangle in which the ∠ACB = the ∠ABC.
We have to prove that AB =AC.
Proof: If AC and AB are equal, suppose that AB>AC.
From BA cut off BD equal to CA, join CD.
Now in the ∆ABC and ∆DBC, AC=BD, BC is common and the included, ∠BCA = the included ∠DBC.
∴ ∆ABC ≌∆DBC [by SAS]
That is, area of ∆ABC = area of ∆DBC. But this is absurd, because a
Part cannot be equal to the whole.
∴ AB is not greater than AC. Similarly AC is not greater than AB.
∴ AB= AC (proved)