Proof.

Let ABC be an isosceles triangle. Extend the equal straight lines AB and AC indefinitely in one direction. Pick a point D at random on the extension of AB. From the extension of AC cut off AE equal in length to AD (proposition I.3). Draw straight lines DC and EB.

Now AD and AC are equal to AE and AB respectively, and angle BAC is common, hence DC and EB are equal, angles ABE and ACD are equal, and angles AEB and ADC are equal (proposition I.4).

Furthermore, BD is equal to CE (common notion 3) and since DC was equal EB together with angle BDC being equal to CEB, the angles DBC and ECB are equal, and angles CBE and BCD are equal (proposition I.4). Thus we have proved our second statement in the proposition.

Since the whole angle ABE was proved equal to angle ACD, and in these angles CBE is equal to angle BCD, then the remaining angle ABC is equal to the remaining angle ACB (common notion 3), and thus the first statement is proven.

Drag e.g. A, B, C, and D.