Given Hilbert's axioms, show that the base angles of an isosceles triangle are congruent.[br][br]Proof: Consider the isosceles triangles [math]\Delta[/math]ABC and [math]\Delta[/math][math]C_1B_1A_1[/math]. Assume that the full SAS condition has been established. Thus, we know that AB [math]\cong[/math] [math]C_1B_1[/math], AC [math]\cong[/math] [math]C_1A_1[/math], and [math]\angle[/math]BAC = [math]\angle[/math][math]B_1C_1A_1[/math]. Note that since the triangles we are considering are isosceles, we can also conclude that BC [math]\cong[/math] [math]B_1A_1[/math]. By Hilbert's Axiom (III-5), we know that [math]\angle[/math]ABC = [math]\angle[/math][math]C_1B_1A_1[/math]. We can then conclude that [math]\angle[/math]ACB = [math]\angle[/math][math]C_1A_1B_1[/math]. Therefore, in triangle [math]\Delta[/math]ABC, the base angles [math]\angle[/math]BAC and [math]\angle[/math]ACB are congruent. Similarly, we know that in triangle [math]\Delta[/math][math]C_1B_1A_1[/math] that base angles [math]\angle[/math][math]B_1C_1A_1[/math] and [math]\angle[/math][math]C_1A_1B_1[/math] are congruent.