$\Delta ABC$ in the figure below:

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

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


Prove: $AB=AC$

enter image description here

The exact version figure should look like: enter image description here

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

  • $\begingroup$ which solution methods are acceptable ? $\endgroup$ – G Cab Aug 15 '17 at 23:31
  • $\begingroup$ I did not find any acceptable solution yet. Do you have any idea? $\endgroup$ – LCFactorization Aug 15 '17 at 23:41
  • 1
    $\begingroup$ Is the method adopted in my answer useful for you? $\endgroup$ – G Cab Aug 18 '17 at 21:31
  • $\begingroup$ Thank you @GCab It is useful to me. Algebraic approach is powerful and universal; Geometric method is simpler and elegant. $\endgroup$ – LCFactorization Aug 19 '17 at 21:12
  • 1
    $\begingroup$ Agreed, with your "tastes" $\endgroup$ – G Cab Aug 20 '17 at 13:42

enter image description here

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 $AB=AC$

  • $\begingroup$ How $\angle 4<\angle 2$ implies $CD<BE$? $\endgroup$ – bigant146 Sep 24 '17 at 18:27
  • 1
    $\begingroup$ 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 $\endgroup$ – 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.


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

  • $\begingroup$ When $D$ is close to $C$, $\angle 2$ is small. It's strange that $\angle 2$ is monotonic... $\endgroup$ – bigant146 Aug 13 '17 at 19:20
  • $\begingroup$ @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.) $\endgroup$ – Brian Chandler Aug 13 '17 at 19:29
  • $\begingroup$ 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 $\endgroup$ – bigant146 Aug 13 '17 at 19:35
  • $\begingroup$ 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) $\endgroup$ – Brian Chandler Aug 13 '17 at 19:52
  • $\begingroup$ 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.) $\endgroup$ – Brian Chandler Aug 13 '17 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.