Isosceles triangle

Question

$\Delta ABC$ in the figure below:

$\angle 1+\angle 2=\angle 3+\angle 4,\quad$

$E\in AB,\; D\in AC,\; F=BD\cap CE,$

$BD=CE$.

Prove: $AB=AC$

The exact version figure should look like:

This problem should be a little more difficult than the Steiner-Lehmus Theorem.

Answer 1

I exchanged the postions of $\angle 3, 4$ in the figure above, which should not affect the result. The proof below will use this new figure.

Construct $EG{/\!\!/}BD$ and $DG{/\!\!/BE}$, then $BDGE$ is a parallelogram.

Connect points $G$ and $C$, label angles $\measuredangle 1=\angle EGD=\angle 1$, $\angle 5 =\angle CGD$ and $\angle 6=\angle DGC$.

Then $CE=BD=EG\Rightarrow \measuredangle 1+\angle 6=\angle 3+\angle 5$

Next use the proof by contradition technique:

If $\angle 3=\angle 1$, the result will be easy to prove, so we suppose $\angle 3\ne\angle 1$. Specifically, without the loss of generality, let us suppose:

$\angle 3>\angle 1\Rightarrow \angle 5<\angle 6\tag{1}$

and $\because\angle 1+\angle 2=\angle 3+\angle 4$

On the other hand, the current assumption leads to：

$\angle 3>\angle 1\Rightarrow \angle 4<\angle 2\Rightarrow CD<BE\Rightarrow CD<DG\Rightarrow \angle 6<\angle 5\tag{2}$

(1) conflicts with (2); similarly, the assumption $\angle 3<\angle 1$ will also lead to such contradiction， therefore, we conclude that:

$\angle 1=\angle 3$ ，

therefore $\angle 2=\angle 4$ ，and $\angle 5=\angle 6$;

and then $\angle 1+\angle 4=\angle 3+\angle 2$ and $AB=AC$

Answer 2

I am going to use the original diagram. (The so-called "exact version" diagram has some labels changed, and I think does not help, because it is too easy to assume the answer from it.)

Given $BD=CE$, we are trying to show that the relation governing the angles: $\angle 1 + \angle 2 = \angle 3 + \angle 4$ implies that $AB=AC$. First, observe that the inverse implication is trivially true:

If $AB=AC$, then $BEDC$ is a trapezium (trapezoid in American) with equal base angles and equal diagonals, and therefore $\angle 1 = \angle 3$ and $\angle 2 = \angle 4$ and the equality follows.

But now consider drawing the two lines from the vertex $A$, and fixing the points $C$ and $E$, while allowing the point $D$ to move along the line $AC$, as the point $B$ similarly moves along the line $AE$ to maintain the equality of lengths $EC$ and $BD$. The $\angle 3$ is fixed: consider how the value of $\angle 1 + \angle 2 - \angle 4$ varies.

If $D$ starts close to $A$, then $\angle 1$ will be arbitrarily small, $\angle 2$ will be arbitrarily small or negative, but $\angle 4$ will be significantly large. The value of $\angle 1 + \angle 2 - \angle 4$ will be very small or negative. As $D$ moves downwards, $\angle 1$ and $\angle 2$ are both monotonically increasing, while $\angle 4$ is monotonically decreasing. Therefore, a fortiori, the value of $\angle 1 + \angle 2 - \angle 4$ is monotonically increasing. There must therefore be a unique point at which the angle relation holds. But we have already shown that it holds if $AB=AC$, and therefore the implication goes both ways, and we have proved that given the angle relation, $AB=AC$. QED

Answer 3

We do not loose generality if we place the triangle with the base in $(-1,0),\;(1,0)$.

Let's call the angles $\angle1, \, \cdots, \, \angle 4$ as $\alpha_1,\, \cdots, \, \alpha_4$ (just to better handle them symbolically), then we must have $$ \bbox[lightyellow] { \alpha _{\,1} + \alpha _{\,2} = \alpha _{\,3} + \alpha _{\,4} \quad \Rightarrow \quad \left\{ \matrix{ \alpha _{\,4} - \alpha _{\,1} = \alpha _{\,2} - \alpha _{\,3} = \delta \hfill \cr \alpha _{\,4} + \alpha _{\,1} = \delta + 2\alpha _{\,1} = \delta + \beta \hfill \cr \alpha _{\,2} + \alpha _{\,3} = \delta + 2\alpha _{\,3} = \delta + \gamma \hfill \cr} \right. } \tag{1}$$

Consider now two segments of length $r$ departing from points $B$ and $C$, and ending at points $$ \bbox[lightyellow] { \eqalign{ & D = \left( { - 1 + r\cos \left( {\delta + \beta /2} \right),\;r\sin \left( {\delta + \beta /2} \right)} \right) \cr & E = \left( {1 - r\cos \left( {\delta + \gamma /2} \right),\;r\sin \left( {\delta + \gamma /2} \right)} \right) \cr} } \tag{2}$$ so that they satisfy the conditions imposed on them for the length, and for the angles (they shall bisect $\beta$ and $\gamma$).

Let's consider then the lines $C,D$ and $B,E$. Their equations are $$ \bbox[lightyellow] { \left\{ \matrix{ {\rm line}\,{\rm CD}:\;{{x - 1} \over { - 2 + r\cos \left( {\delta + \beta /2} \right)}} = {y \over {r\sin \left( {\delta + \beta /2} \right)}} \hfill \cr {\rm line}\,{\rm BE}:{{x + 1} \over {2 - r\cos \left( {\delta + \gamma /2} \right)}} = {y \over {r\sin \left( {\delta + \gamma /2} \right)}} \hfill \cr} \right. } \tag{3}$$ and we want their slopes to be: $$ \bbox[lightyellow] { \eqalign{ & \left\{ \matrix{ {{r\sin \left( {\delta + \beta /2} \right)} \over { - 2 + r\cos \left( {\delta + \beta /2} \right)}} = - \tan \left( {\delta + \gamma } \right) \hfill \cr {{r\sin \left( {\delta + \gamma /2} \right)} \over {2 - r\cos \left( {\delta + \gamma /2} \right)}} = \tan \left( {\delta + \beta } \right) \hfill \cr} \right.\quad \Rightarrow \quad (4.a) \cr & \Rightarrow \quad \left\{ \matrix{ {{\sin \left( {\delta + \beta /2} \right)} \over {2/r - \cos \left( {\delta + \beta /2} \right)}} = \tan \left( {\delta + \gamma } \right) \hfill \cr {{\sin \left( {\delta + \gamma /2} \right)} \over {2/r - \cos \left( {\delta + \gamma /2} \right)}} = \tan \left( {\delta + \beta } \right) \hfill \cr} \right.\quad \Rightarrow \quad (4.b) \cr & \Rightarrow \quad \left\{ \matrix{ \sin \left( {2\delta + \beta /2 + \gamma } \right) = 2/r\sin \left( {\delta + \gamma } \right) \hfill \cr \sin \left( {2\delta + \beta + \gamma /2} \right) = 2/r\sin \left( {\delta + \beta } \right) \hfill \cr} \right. \quad (4.c) \cr} }$$

The system of equations in (4.c) above can be represented as $$ \bbox[lightyellow] { \left\{ \matrix{ 0 \le \beta ,\gamma < \pi /2 - \delta \hfill \cr F\left( {\beta ,\;\,\gamma \;;\;\,\delta ,r} \right) = \sin \left( {2\delta + \beta /2 + \gamma } \right) - 2/r\sin \left( {\delta + \gamma } \right) \hfill \cr F\left( {\beta ,\;\,\gamma \;;\;\,\delta ,r} \right) = 0 \hfill \cr F\left( {\gamma ,\;\beta \;\,;\,\;\delta ,r} \right) = 0 \hfill \cr} \right. } \tag{5}$$ and since it imposes to be null either the $F(\beta,\,\gamma)$ and its symmetric $F(\gamma,\, \beta)$, then, clearly, if there are, the solutions will be $\beta=\gamma$, i.e. the triangle must be isosceles.
Q.E.D.