Visual proof of isosceles base-angle congruency?

Question

A geometric proof (without algebra or trigonometry), and ideally presented visually (a proof without words). EDITED

(I'm especially curious if it's possible without without using triangle congruency.)

1.One proof draws a line from the vertex and bisects the base. The two triangles formed have side-side-side congruency (the bisected sides are equal, the common side is equal, and the remaining sides are equal because isoaceles legs), and the base angles are equal because corresponding angles.

2. A similar approach instead drops from then vertex a perpendicular of the base. Now we have angle-side-side congruency (perpendicular right angle, common side, isosoceles legs), and again the base angles are corresponding.

Is there a simpler proof?

I was hoping for one using only parallels, and alternate interior and corresponding angles, but these couldn't use the equal lengths of the legs (could they?)

3. EDIT There's an amazing one here

Consider the triangles ΔABC, ΔACB. These are congruent via SSS criterion and hence ∠ABC = ∠ACB.

4. Blue's comment notes it's also congruent by SAS.

5. EDIT2 There's also Euclid's Elememts I, 5 (from mweiss's comment).

All these methods show angle congruence via triangle congruence. Is there any other way of using lengths to show something?

e.g. angles congruency can be shown without triangle congruence, e.g. vertical angles, corresponding angles. Are there ways that lengths can play a role, apart from via triangle comgriency?

Answer 1

Let $AB$ and $AC$ be the equal sides of $\triangle ABC$, construct AD is perpendicular to BC.

from $\triangle ABD $ and $\triangle ACD $,

$AB=AC$

$AD$ is common

$\angle ADB= \angle ADC$, since both are right angles.

by $SSA $ congruence

$\triangle ABD \cong \triangle ACD$

therefore base angles are equal.

Answer 2

use sine rule, $$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$$, by definition of isosceles triangle, any two sides are equal.

w.l.o.g. let, $a=b$ $$\frac{a}{sinA}=\frac{a}{sinB}=\frac{c}{sinC}$$ $$\frac{a}{sinA}=\frac{a}{sinB}$$ $$\frac{a}{a}=1=\frac{sinA}{sinB}$$ $$\implies sinA=sinB $$ and $\angle A $ and $\angle B $ are less than $180^o$.

therefore $\angle A =\angle B $

Answer 3

HINT:

use cosine rule $$a^2=b^2+c^2-2bc.cosA$$

Solution

w.l.o.g. $a=b$,

substitute $a=b$ ,we get $$a^2=a^2+c^2-2ac.cosA$$ $$a^2=a^2+c^2-2ac.cosB$$ equating the two equations,after substition

$cosA=cosB$

and ∠A and ∠B are less than $180^o$.

therefore ∠A=∠B