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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.

  1. 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?)

  1. EDIT There's an amazing one here

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

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

  2. 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?

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    $\begingroup$ As you mention that I mention, there's the one-liner: $\triangle ABC \cong\triangle ACB$ by SAS; therefore $\angle B\cong\angle C$. Since SAS is usually taken as an axiom (and SSS is a later-derived theorem), it would appear that no proof could be objectively simpler. Now, that argument is a little weird to newbies, so maybe you're asking whether there's a subjectively-"best" proof that satisfies beginners. Ignoring that opinion-based questions are discouraged on Math.SE, I'll note that there's an interesting historical context to this. (continued) $\endgroup$ – Blue May 13 '18 at 5:46
  • $\begingroup$ (continuing) This result earned the name pons asinorum ---Latin for "bridge of asses"--- partly because Euclid's construction resembles a bridge, and partly because this was considered the place where math students transitioned into advanced thinkers. So, historically, this result has been deemed non-obvious enough as to identify the mathematically adept; centuries of instructors seem to have decided that there is no proof that satisfies beginners. (To me, this over-estimates the result's difficulty, and underestimates the student.) $\endgroup$ – Blue May 13 '18 at 6:05
  • $\begingroup$ @Blue I think the first two proofs I gave satisfies beginners (or maybe not 100% "beginners", but later in high school?). After further thought, the confusing thing about the $SAS$ proof may be related to whether $\angle BAC$ and $\angle CAB$ are the same angle or not. And then, that congruency is not saying two things are the same, but whether they have exactly the specified correspondence or not... it seems to me that isosceles and equilateral triangles are the only triangles for which this issue comes up, so it comes across more as a special case than a general rule. $\endgroup$ – hyperpallium May 13 '18 at 7:12
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Let $AB$ and $AC$ be the equal sides of $\triangle ABC$, construct AD is perpendicular to BC.

isosceles with perpendicular

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.

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  • $\begingroup$ Thanks, this is a nice full version of the sketch in Q's second paragraph. BTW images are very helpful in geometry, but it didn't come through. $\endgroup$ – hyperpallium May 19 '18 at 6:08
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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 $

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  • $\begingroup$ Thanks, I didn't stress this in the auestion, but I'm looking for a geometric proof (like your first one), and ideally, presented as a visual proof (like this (Although I used the visualization tag) $\endgroup$ – hyperpallium May 21 '18 at 9:24
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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

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