# 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)