# Isosceles triangle

$\Delta ABC$ 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.

• which solution methods are acceptable ? – G Cab Aug 15 '17 at 23:31
• I did not find any acceptable solution yet. Do you have any idea? – LCFactorization Aug 15 '17 at 23:41
• Is the method adopted in my answer useful for you? – G Cab Aug 18 '17 at 21:31
• Thank you @GCab It is useful to me. Algebraic approach is powerful and universal; Geometric method is simpler and elegant. – LCFactorization Aug 19 '17 at 21:12
• Agreed, with your "tastes" – G Cab Aug 20 '17 at 13:42

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

(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 $$AB=AC$$

• How $\angle 4<\angle 2$ implies $CD<BE$? – bigant146 Sep 24 '17 at 18:27
• Consider the two triangles $BCE$ and $CBD$ have two equal-length sides, then the angle between the two corresponding sides determines the third side. You can use Cosine theorem since both $\angle 2$ and $\angle 4$ are acute angles – user6043040 Sep 26 '17 at 8:34

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.

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

• When $D$ is close to $C$, $\angle 2$ is small. It's strange that $\angle 2$ is monotonic... – bigant146 Aug 13 '17 at 19:20
• @bigant146 : You have only given a half-sentence, but it's a hint. OK, when D is close to C, angle 4 is very small, and both of the other angles are larger than they were when D was at the isosceles position; ergo the angle expression is much greater than angle 3. This is in line with expectations, since the expression is monotonically increasing. (I'm still baffled by the down votes.) – Brian Chandler Aug 13 '17 at 19:29
• In fact, it's only one strange thing in your solution. The function you defined looks strange too: for a fixed point $D$ sometimes there are two possible choices of $B$. That's why $\angle 1$, $\angle 2$ and $\angle 4$ are not functions in general sense – bigant146 Aug 13 '17 at 19:35
• Well, I am probably using slightly sloppy language. I've talked about "moving" D, when really I should have a variable point d, and a variable b for the point on AE, with the length bd equal to EC. Then unquestionably, the position of b determines all three of the angles 1, 2, and 4, whose values are thus a function of the position of b. I should really though have d vary from the position of A to the position where Ad=Ab, to avoid the the ambiguous b problem. (sorry, can't seem to do mathjax here) – Brian Chandler Aug 13 '17 at 19:52
• Sorry, now I see your original complete comment. If the line segment has d moving down from A, then if the fixed angle AEC is obtuse, b will have passed below E; when d reaches C, Ecb is an isosceles triangle. (E would be the "other" position for b, but it can't get there this way!) b has moved always downwards, so angle bCA can only increase. (If angle AEC is acute, it doesn't work; but I think it is easy to demonstrate that is must be obtuse, or the diagram cannot be drawn.) – Brian Chandler Aug 13 '17 at 20:06