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For reference:

My progress:

\begin{align*} & AH \perp FD \\ & \triangle AFD \text{ is isosceles} \quad \therefore \measuredangle BFA = \measuredangle FDA = x \\ & AF = FD \\ & \measuredangle HBA = 180-135^\circ = 45^\circ \quad \therefore \triangle HBA \text{ is isosceles} \end{align*}

I drew some auxiliary lines but it wasn't enough to reach the solution.

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  • $\begingroup$ Answer: $18^o30'$ $\endgroup$ Aug 6, 2021 at 1:43
  • $\begingroup$ Is that an exact answer, I am getting $\cot^{-1}(3)$, which is somewhat close to $18.5^\circ$. $\endgroup$ Aug 6, 2021 at 2:36
  • $\begingroup$ Would someone solve by geometry? $\endgroup$ Aug 6, 2021 at 14:54

4 Answers 4

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Build a parallelogram $ABDE$. Then $$\angle DEA = \angle ABD = 135^\circ = \angle DCA,$$ hence $ADEC$ is cyclic. Moreover, $\angle EAD = \angle BDA = x$, hence $$\angle CAE = \angle CAD - \angle EAD = 2x-x=x.$$ As a consequence, $CE=DE$ because the angles subtended by arcs $CE$, $DE$ of the red circle are equal.

Now, it is easy to calculate that angle $CBD$ equals $90^\circ + x$ and that the concave angle $CED$ equals $180^\circ + 2x$. Since $CE=DE$, it follows that $B$ lies on the circle with center $E$ and radius $CE=DE$. Hence $\angle DEB = 2\angle DCB = 90^\circ$. Since $DE=BE$, the right triangle $DEB$ is isosceles.

Let $F$ be the midpoint of $BE$. Let $G$ be the midpoint of $BD$ and $H$ be the midpoint of $BG$. Easy to see that triangles $GFB$ and $GFH$ are isosceles right triangles, hence $HF = HB = \frac 12BG = \frac 14 BD$, so $HD = BD - BH = 4HF - HF = 3HF$. This yields $\tan x = \frac{HF}{HD} = \frac 13$ so the answer is $$x = \arctan \frac 13.$$

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I have a trigonometry solution. Looking at the form of the answer, I don't think there will exist a purely synthetic approach.

With some angle chasing, we get that $$\angle ADC=45-2x$$ $$\angle ABC=45-x$$ Using law of sines on $\Delta ACD$, we get $$\frac{AC}{\sin (45-2x)}=\frac{AD}{\sin 135}$$ Using law of sines on $\Delta ABD$, we get $$\frac{AB}{\sin (x)}=\frac{AD}{\sin 135}$$ Combining these two equations gives $$\frac{AC}{\sin (45-2x)}=\frac{AB}{\sin (x)}$$ $$\frac{AC}{AB}=\frac{\sin (45-2x)}{\sin (x)}$$ With simple trig definitions on right triangle $\Delta ABC$, we get that $$\sin (45-x)=\frac{AC}{AB}$$ Hence, $$\frac{\sin (45-2x)}{\sin (x)}=\sin (45-x)$$ Using angle sum/difference identities along with others, this simplifies to $$\frac{\sqrt{2}}{2}(\cos (2x)-\sin (2x))=\frac{\sqrt{2}}{2}(\sin (x)\cos (x)-\sin^2 x)$$ $$\cos^2 (x)-\sin^2 (x)-2\sin (x)\cos (x)=\sin (x)\cos (x)-\sin^2 (x)$$ $$\cos^2 (x)=3\sin (x)\cos (x)$$ Note that if $\cos (x)=0$, then $x=\frac{\pi}{2}$. However, it is clear by the diagram that $x$ is acute (we can get even better bounds, but it is unnecessary).

Hence, we have $\cos (x)\neq 0\implies$ $$\cos (x)=3\sin (x)$$ $$x=\boxed{\cot^{-1} (3)}$$

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  • $\begingroup$ I believe there must be a geometric solution drawing auxiliary lines... the exercise is from a geometry book $\endgroup$ Aug 6, 2021 at 11:25
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enter image description here

Hints: You can easily see that:

$2x+\widehat {ADC}=45^o$

Now try to find another relation between these two angles using data the figures give.

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  • $\begingroup$ would be FG or FC? $\endgroup$ Aug 6, 2021 at 19:26
  • $\begingroup$ @petaarantes, FC is correct . I corrected it. $\endgroup$
    – sirous
    Aug 6, 2021 at 19:53
  • $\begingroup$ @sirous..sorry i sorry, i couldn't find this relation $\endgroup$ Aug 8, 2021 at 1:18
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Once again, Pythagoras theorem does the job.

enter image description here

Say, $BC = a, AC = b$. Drop a perp from $B$ to $AD$. Notice that right triangles $\triangle ABC \cong \triangle ABH$, given $\angle ABC = \angle BAH = 45^\circ - x$ and $AB~$ is hypotenuse to both triangles.

So, $AH = BC = a, BH = AC = b$

Similarly notice that $\triangle DBE \cong \triangle BDH$. So, $DE = BH = b, DH = BE$ but $BE = CF = DF = a + b$

That leads to $DF = a + b, AF = a + 2b, AD = 2a + b$

Applying Pythagoras,
$(a+b)^2 + (a+2b)^2 = (2a+b)^2 \implies (a + b) (a-2b) = 0$

So we get $a = 2b$ and that shows $\triangle ADF$ is $3:4:5$ triangle.

Hence, $2x \approx 37^\circ$

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  • $\begingroup$ excellent...great..thansk $\endgroup$ Jan 25 at 10:42

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