What is $\angle AEB$?

Question

Let $E$ be a point inside the square $ABCD$. and $|EC|=3,|EA|=1,|EB|=2$

What is the angle $\widehat {AEB}$?

I can only find $|ED|=\sqrt{12}$

Answer 1

Here is a less algebraicly intensive solution and only requires knowledge of the sine law and the cosine law, and solving quadratics of the form $x^2 = a$(super easy) all of which is material known by the average grade 10 math student in Canada. From the sine law we know that
\begin{align} \frac{\sin(ABE)}{1} = \frac{\sin(AEB)}{s} && \frac{\sin(CBE)}{3} =\frac{\sin(90 -ABE)}{3} = \frac{\cos(ABE)}{3} = \frac{\sin(BEC)}{s} \end{align} After re arranging and squaring both parts we get \begin{align} \sin^2(ABE) = \frac{\sin^2(AEB)}{s^2} && \cos^2(ABE) = \frac{9\sin^2(BEC)}{s^2} \end{align} Recall that $\cos^2(x) = 1 - \sin^2(x)$. \begin{align} \frac{9\sin^2(BEC)}{s^2} &= 1 - \frac{\sin^2(AEB)}{s^2}\\ 9(1 - \cos^2(BEC)) &= s^2 - 1 + \cos^2(AEB) \end{align} Using the cosine law we know that $5 - 4\cos(AEB) = s^2$ and $13 - 12\cos(BEC) = s^2$. Setting the equations equal and isolating for $\cos(BEC)$ yields $\cos(BEC) = \frac{\cos(AEB)}{3} + \frac{2}{3}$. We can know solve our equation \begin{align} 9(1 - (\frac{\cos(AEB)}{3} + \frac{2}{3})^2) &= 5 - 4\cos(AEB) - 1 + \cos^2(AEB)\\ 9(1 - \frac{\cos^2(AEB)}{9} - \frac{4\cos(AEB)}{9} - \frac{4}{9}) &= 4 - 4\cos(AEB) + \cos^2(AEB)\\ 5 - \cos^2(AEB) - 4\cos(AEB) &= 4 - 4\cos(AEB) + \cos^2(AEB)\\ \end{align} And since the $4\cos(AEB)$ terms cancel out we get $2\cos^2(AEB) - 1 = 0$ yielding \begin{equation} \cos(AEB) = \frac{\pm 1}{\sqrt{2}} \iff \angle AEB = 45, 135 \end{equation} And since $AEB$ can't be 45 $\angle AEB = 135 \deg$.

Answer 2

Let $E^\prime$ be the reflection of $E$ through $AB$, $E^{\prime\prime}$ the reflection of $E$ through $BD$. It follows that

1. $AE^{\prime\prime}=CE=3$.
2. $BE^\prime=BE^{\prime\prime}=BD=2$.
3. $AE^\prime=AE=1$.
4. $\angle CBE^{\prime\prime}=\angle E^\prime BA=\angle EBA$.

It follows further that $\angle E^\prime BE^{\prime\prime}=90^\circ$. Thus we have $$E^\prime E^{\prime\prime}=BE^\prime\sqrt{2}=2\sqrt{2}, \angle BE^\prime E^{\prime\prime}=45^\circ\tag{5}.$$ So in $\triangle AE^\prime E^{\prime\prime}$ we have $${AE^\prime}^2+{E^\prime E^{\prime\prime}}^2={AE^{\prime\prime}}^2,$$ or equivalently $\angle AE^\prime E^{\prime\prime}=90^\circ$. Together with (5), we have $\angle BEA=\angle BE^\prime A=135^\circ$.

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Btw, I think that this is way too difficult for a multiple choice question.

Answer 3

Take $F$ outside of square such that $\bigtriangleup AEB \equiv \bigtriangleup CFB$. Then $EF = 2\sqrt2$, that means $\angle CFE = 90^{\circ}$. $\angle AEB = \angle CFA = 90+45^{\circ}$.

Answer 4

There has to be an easier way, but this works...

Use the law of cosines to get:

$s^2 = 5 - 4\cos(AEB)$

$s^2 = 13 - 12\cos(BEC)$

$2s^2 = 10 - 6\cos(AEB+BEC)$

Eliminate $s^2$, and use $\cos(AEB + BEC) = \cos(AEB)\cos(BEC) - \sin(AEB)\sin(BEC)$. Exchange the sines for cosines, eliminate $\angle BEC$, and a whole bunch of algebra later we get:

$(4\cos(AEB) - 5)(2\cos^2(AEB) - 1) = 0$

Reject the first root for being too large, and we get $\cos(AEB) = \pm \frac{1}{\sqrt{2}}$.

Reject $45^\circ$ because that implies that $s = \sqrt{5 - \frac{4}{\sqrt{2}}} \approx 1.473 < \frac{3}{\sqrt{2}}$, so $E$ would not be in the square. We can similarly reject $225^\circ$ and $315^\circ$.

Therefore, $\angle AEB = 135^\circ$.

Answer 5

Take a horizontal reflection of the square $ \ \square ABCD \ $ of side $ \ s \ $ and its interior segments as specified in the problem and superimpose it on the original square. We then have two isosceles trapezoids $ \ EE'CB \ $ and $ \ EE'DA \ $ with parallel sides of length $ \ u \ $ (of the segment $ \ EE' \ ) \ $ and $ \ s \ $ and known slant heights; we also know the lengths of the diagonals of trapezoid $ \ EE'CB \ \ . \ $ We let $ \ h \ $ be the height of $ \ EE'CB \ \ ; \ $ the height of $ \ EE'DA \ $ is then $ \ s - h \ \ . \ $ The line $ \ EE' \ $ extends to intersect the sides of the square at $ \ F \ $ and $ \ H \ \ : \ $ the lengths of $ \ EF \ $ and $ \ E'H \ $ are then $ \ \frac{s \ - \ u}{2} \ \ , \ $ which segments are also the altitudes of congruent triangles $ \ \triangle AEB \ $ and $ \ \triangle DE'C \ \ . $ The altitudes of the trapezoids at $ \ E \ $ meet the sides of the square at $ \ G \ $ and $ \ I \ \ ; \ $ the segment $ \ CG \ $ thus has length $ \ \frac{s \ + \ u}{2} \ \ . \ $ Lastly, we call $ \ \theta \ $ the sought measure of $ \ \angle AEB \ \ . $

The "Pythagorean Theorem" permits us to write $$ (s - h)^2 \ + \ \left(\frac{s \ - \ u}{2} \right)^2 \ \ = \ \ 1^2 \ \ \ , \ \ \ h^2 \ + \ \left(\frac{s \ - \ u}{2} \right)^2 \ \ = \ \ 2^2 \ \ \ , \ \ \ h^2 \ + \ \left(\frac{s \ + \ u}{2} \right)^2 \ \ = \ \ 3^2 \ \ \ . $$ Subtracting the second of these equations from the third yields $$ \left(\frac{s \ + \ u}{2} \right)^2 \ - \ \left(\frac{s \ - \ u}{2} \right)^2 \ \ = \ \ 9 \ - \ 4 \ \ \Rightarrow \ \ su \ = \ 5 \ \ . $$

The triangles $ \ \triangle AEB \ $ and $ \ \triangle DE'C \ $ have areas $ \ \mathcal{A}_{\Delta} \ = \ \frac12·s·\left(\frac{s \ - \ u}{2} \right) \ = \ \frac14· ( s^2 \ - \ su ) \ \ , \ $ but also for "included angle" $ \ \angle AEB \ \ , \ \ \mathcal{A}_{\Delta} \ = \ \frac12·1·2·\sin \theta \ = \ \sin \theta \ \ . \ $ (It may also be remarked that, by the triangle inequality, $ \ 2 \ < \ s \ < \ 3 \ \ . \ ) $

The area of $ \ \square ABCD \ $ is the sum of

$$ \mathcal{A}_{\square} \ \ = \ \ s^2 \ \ = \ \ \mathcal{A}_{EE'CB} \ + \ \mathcal{A}_{EE'DA} \ + \ 2·\mathcal{A}_{\Delta} $$ $$ = \ \ \left(\frac{u \ + \ s}{2} \right)·h \ + \ \left(\frac{u \ + \ s}{2} \right)·(s \ - \ h) \ + \ 2·\sin \theta \ \ = \ \ \frac12 s^2 \ + \ \frac12 su \ + \ 2·\sin \theta $$ $$ \Rightarrow \ \ \frac12 s^2 \ \ = \ \ \frac52 \ + \ 2·\sin \theta \ \ \Rightarrow \ \ s^2 \ \ = \ \ 5 \ + \ 4·\sin \theta \ \ . $$

We also have, applying the Law of Cosines to $ \ \triangle AEB \ \ , $ $$ s^2 \ \ = \ \ 1^2 \ + \ 2^2 \ - \ 2·1·2·\cos \theta \ \ = \ \ 5 \ - \ 4 · \cos \theta \ \ . $$

Consequently, $$ 5 \ + \ 4·\sin \theta \ \ = \ \ 5 \ - \ 4 · \cos \theta \ \ \Rightarrow \ \ \sin \theta \ \ = \ \ -\cos \theta \ \ \Rightarrow \ \ \tan \theta \ \ = \ \ -1 \ \ , $$ which in a triangle can only mean $ \ \boxed{ \ \theta \ = \ 135º \ } \ \ . $

From what we've found, it becomes clear that we would not wish to work with the value of $ \ s \ $ directly, say, in any trigonometric approach, as $ \ s^2 \ = \ 5 + 4·\frac{\sqrt2}{2} \ \Rightarrow \ s \ = \ \sqrt{5 + 2 \sqrt2} \ \approx \ 2.798 \ $ has no simpler expression. [We can verify, incidentally, that $$ \ \mathcal{A}_{\Delta} \ \ = \ \ \frac14· ( s^2 \ - \ su ) \ \ = \ \ \frac14· ( 5 + 4·\frac{\sqrt2}{2} \ - \ 5 ) \ \ = \ \ \frac{\sqrt2}{2} \ \ = \ \ \sin \theta \ \ . \ ] $$

This is a rather interesting geometry problem in that it seems to have so many approaches to solution (see also the "Linked" post).

Answer 6

I had asked a similar question, without knowing this was already asked, so I'll post the same answer I posted there, here as well:

This is my approach:

Rotate $\triangle AEC$ counterclockwise by $90^\circ$ about point $C$ and form $\triangle CFD$. Since $CE=CF=2$, and $\angle ECF=90^\circ$, we know that $EF=2\sqrt{2}$, and that $\angle CEF=\angle CFE=45^\circ$. Notice that the sides of $\triangle EFD$ are Pythagorean Triples $(3^2=(2\sqrt{2})^2+1^2)$, that means $\angle EFD=90^\circ$.

Therefore $x=45^\circ+90^\circ=135^\circ$