# Calculate angle $x$ in the figure

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.

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

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 $$x = \arctan \frac 13.$$

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

• I believe there must be a geometric solution drawing auxiliary lines... the exercise is from a geometry book Aug 6, 2021 at 11:25

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.

• would be FG or FC? Aug 6, 2021 at 19:26
• @petaarantes, FC is correct . I corrected it. Aug 6, 2021 at 19:53
• @sirous..sorry i sorry, i couldn't find this relation Aug 8, 2021 at 1:18

Once again, Pythagoras theorem does the job.

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

• excellent...great..thansk Jan 25 at 10:42