In an equilateral triangle $ ABC $ the point $ D $ and $ E $ are on sides $ AC $ and $AB$ respectively, such that $ BD $ and $ CE $ intersect at $P$ , and the area of the quadrilateral $ ADPE $ is equal to area of $ \Delta BPC $ find $ \angle BPE $. This question when I first tried looked easy and I was also able to guess the answer but when I tried to proof, was not able to work it out. I want some help. Thank you. No image was provided in question. I am attaching my drawing. image

  • $\begingroup$ Could you please provide an image? $\endgroup$ – Dr. Mathva Apr 9 at 16:08
  • $\begingroup$ It might be that all points P with the above property generate the same angle BPE, but I think that this is hard question. Are there no further restrictions on P, e.g. P is lying on symmetry axis of BC? $\endgroup$ – Daniel W. Apr 9 at 17:23
  • $\begingroup$ No there were no further restrictions neither any other specification. But after some trial I guessed that answer should be 60° but unfortunately didn't get a way to show or prove it $\endgroup$ – Keshav Sharma Apr 9 at 17:43
  • $\begingroup$ I think, the key is to show it in the case were P is on the symmetry axis of BC and then show that all points P with this property lie on a circle through B and C to use the Theorem of the circumferential angle. $\endgroup$ – Daniel W. Apr 9 at 18:00
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    $\begingroup$ Welcome to Math.SE! Please use MathJax. For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 10 at 11:11

From the condition, we add area of ∆BPC to both sides of the equation, this becomes

Area of ∆AEC = Area of ∆CDB

It is then easy to prove that ∆AEC is congruent to ∆CDB. So ∠BDC = ∠CEA.

Thus, quadrilateral ADPE is concyclic.

so ∠BPE = ∠BAC = 60˚

  • $\begingroup$ Nice and easy solution! +1 $\endgroup$ – Michael Rozenberg Apr 9 at 22:20
  • $\begingroup$ As I said, this is elegance itself! But I think that you wanted to add the area of $\triangle DPC$ to both sides of the equation in your 1st sentence, and not $\triangle BPC.$ I have now understood what is going on. There are indeed infinitely many configurations satisfying the data, as I'd deduced, but the remarkable thing is that -- as you have proved -- they all always have $\angle BPE=60°.$ I think this is a beautiful result. I had been assuming that we should have $\angle BPE=60°$ only in the case where $AE=EB$ because I'd been trying to determine $|AE|,$ say, satisfying the conditions. $\endgroup$ – Allawonder Apr 11 at 9:01

The hint.

Prove that $$\left(S_{\Delta BPC}\right)^2=S_{\Delta APB}S_{\Delta APC}.$$ I got $$\measuredangle BPE=60^{\circ}.$$

Let $S_{\Delta BPC}=a$, $S_{\Delta PAC}=b$ and $S_{\Delta PAB}=c$.

Thus, $$\frac{S_{\Delta PEB}}{c}=\frac{BE}{AB}=\frac{BE}{BE+EA}=\frac{1}{1+\frac{EA}{BE}}=\frac{1}{1+\frac{b}{a}}=\frac{a}{a+b},$$ which gives $$S_{\Delta PEB}=\frac{ac}{a+b}.$$ Similarly, $$S_{\Delta PDC}=\frac{ab}{a+c}.$$ Thus, $$S_{AEPD}=b+c-\frac{ab}{a+c}-\frac{ac}{a+b}=\frac{bc}{a+c}+\frac{bc}{a+b}.$$ Id est, $$\frac{bc}{a+c}+\frac{bc}{a+b}=a$$ or $$a^2=bc$$ or $$\frac{a}{b}=\frac{c}{a}$$ or $$\frac{BE}{AE}=\frac{AD}{CD},$$ which gives $$DC=AE,$$ $$\Delta AEC\cong\Delta CDB,$$ which gives $$\measuredangle BPE=\measuredangle DBC+\measuredangle ECB=\measuredangle DBC+60^{\circ}-\measuredangle ACE=60^{\circ}.$$

  • $\begingroup$ What does the notation $S_{\triangle \cdots}$ mean? $\endgroup$ – Allawonder Apr 11 at 8:56
  • $\begingroup$ @Allawonder $S_{\Delta ABC}$ it's an area of the triangle ABC. $\endgroup$ – Michael Rozenberg Apr 11 at 9:10
  • $\begingroup$ Alright. So how did you derive the first equality? $\endgroup$ – Allawonder Apr 11 at 9:17
  • $\begingroup$ @Allawonder My solution is not so nice and I prefer don't post it. The best solution it's the qsmy's solution. By the way, if you still want to see my solution I can show. $\endgroup$ – Michael Rozenberg Apr 11 at 9:51
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    $\begingroup$ @Allawonder I added something. See now. $\endgroup$ – Michael Rozenberg Apr 11 at 21:05

enter image description here

Let $a$ be the side length of the equilateral $\triangle ABC$, $\angle BPE=\theta$, $\angle PBC=\phi$, and let $[\cdot]$ denote the area. Note that

\begin{align} [AEPD]&=[ABC]-[BCD]-[BCE]+[BCP] , \end{align}

and since $[AEPD]=[BCP]$, we must have \begin{align} [BCD]+[BCE]&=[ABC] . \end{align}

Using the three-angles-and-a-side expression for the area, we get \begin{align} \frac{\sqrt3}4\cdot a^2\cdot \frac{\sin(\phi)}{\sin(120^\circ-\phi)} + \frac{\sqrt3}4\cdot a^2\cdot \frac{\sin(\theta-\phi)}{\sin(120^\circ+\phi-\theta)} &= \frac{\sqrt3}4\cdot a^2 ,\\ \frac{\sin(\phi)}{\sin(120^\circ-\phi)} + \cdot \frac{\sin(\theta-\phi)}{\sin(120^\circ+\phi-\theta)} -1 &=0 . \end{align}

After expanding and refactoring we get

\begin{align} \frac{(\sin^2\phi+\cos^2\phi)(\sqrt3\sin\theta-3\cos\theta)}{ (\sqrt3\cos\phi+\sin\phi) (\sqrt3\cos\theta\cos\phi +\sqrt3\sin\theta\sin\phi+\sin\theta\cos\phi -\cos\theta\sin\phi) } &=0 , \end{align}

and it follows that $\theta=60^\circ$.


IF there exists a set of points P for which the two areas are equal, then one such point P lies on the vertical altitude (axis of symmetry), thus we only need to find that point. Dividing each of the two areas in half with the vertical altitude, the problem can be stated as Area(BPS) = Area (APE), where S is the foot of the altitude (the midpoint of BC). The midpoint (centroid?) of the triangle (with altitudes) divides the area of the whole triangle into six similar triangles, including the two above. So the midpoint of the triangle satisfies the condition, thus the angle is 60 degrees.


Let the areas of the each of the four parts, the quadrilateral, $\triangle EBP,\triangle BCP,$ and $\triangle PCD$ respectively be $W,X,Y,Z.$ Also, letting $|AE|=a,|EB|=b,|AD|=m,|DC|=n,$ we have that (considering the two triangles based on $AB$): $$\frac{W+Z}{X+Y}=\frac ab.$$ Similarly, by considering the triangles based on $AC,$ we must have $$\frac{W+X}{Z+Y}=\frac mn.$$ Since we want $W=Y,$ it follows that we must have $$\frac ab=\frac nm.$$ We must also have $$a+b=n+m=1,$$ that last equality holding wlog. We then have that $a=n$ and $b=m,$ from the first two equations. It then becomes clear that there are infinitely many positive $a,b,m,n$ satisfying the conditions given. If we let, in addition, $a=b$ (which is not actually given), then it is clear that in this case there is a properly defined solution. Indeed, $ED$ now becomes parallel to $BC,$ and $\triangle AED$ is similar to $\triangle ABC.$ Thus, we have (as is easily verified) that $\angle BPE=60°.$ However, I emphasise that (as OOP confirmed under the comments on OP) the assumed condition $a=b$ is not part of the original problem. Hence my claim above that the problem has no unique solution. However, we may get a parametric solution as follows.

Fix any $0<a<1,$ then the position of the cevian $CE$ is fixed. Also, as is clear from the above argument, $m,$ and therefore the other cevian $BD,$ is fixed too. Thus, the angle $BPE$ sought is also now determined. Therefore, let $\angle BPE=\angle DPC=\theta.$ Let us make the following abbreviations, letting $\angle PEB=\omega,\angle PBC=\phi,\angle PCB=\phi',\angle CDP=\omega'.$

Now, applying the sine rule in $\triangle EBC,$ we obtain $$\frac {1}{\sin\omega}=\frac{b}{\sin\phi'},$$ so that $\sin\phi'=b\sin\omega.$ But in the same triangle, we have $\omega+60°+\phi'=180°,$ upon substitution of which in the previous equation and some other transformations, gives $$\cos(30°-\omega)=(1-a)\sin\omega.$$ Performing similar operations in the triangle $DBC$ gives the equation $$\sin\phi=m\cos(30°-\phi).$$ Thus, the angles $\phi$ and $\omega$ are now explicitly known. Finally, in $\triangle EBP,$ we have $\omega+(60°-\phi)+\theta=180°,$ which gives the angle sought $$\theta=120°+\phi-\omega.$$ This is only determined uniquely once we select any real number $a$ such that $$0<a<1.$$

PS. It should be noted that the procedure above will apply mutatis mutandis to an arbitrary triangle with given sides, since then the angles are also determined by the cosine rule, and therefore the data in this case is not less than that given in the equilateral case above.


In light of the elegant proof of @qsmy I was wondering where I had gone wrong. Turns out both his results and mine are consistent, provided I withdraw the non-sequitur that because there are infinitely many arrangements satisfying the given conditions, it follows that there are also infinitely many solutions. It does not actually follow. That was my error. As qsmy has shown, the angle $BPE$ is always equal to $60°$ for all those configurations. Isn't that cool!

  • $\begingroup$ How does the equality confirms the claim that there are infinitely many solutions? and please see the other answers $\endgroup$ – qsmy Apr 9 at 21:56
  • $\begingroup$ @qsmy The other answers indeed agree that the angle is $60°.$ I shall need to check my solution again. $\endgroup$ – Allawonder Apr 10 at 19:26
  • $\begingroup$ @qsmy I have now confirmed that there is indeed no unique solution to the problem as presented in OP. Also, I have rewritten my answer to make my argument clearer. $\endgroup$ – Allawonder Apr 10 at 22:44
  • $\begingroup$ I would say whatever your deduction is is true even if the condition 0<a<1 does not exist. $\endgroup$ – qsmy Apr 10 at 23:41
  • $\begingroup$ @qsmy I don't understand. Are you saying that you have not found a fault in my reasoning? If so, why does it seem like my result (that the solution is not unique) seems to contradict yours, for example, which gives a unique solution? Something's got to be wrong somewhere. $\endgroup$ – Allawonder Apr 10 at 23:45

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