# I want to determine the lenght of $x$ by making equations for each triangle ACD and ADB.

I want to determine the lenght of $$x$$ by making equations for each triangle ACD and ADB.

Let's recall $$\angle DAB = \alpha$$, $$\angle DBA = \beta$$ in triangle ADB

$$\alpha + \beta + 90 = 180$$

$$\alpha + \beta =180 \tag{1}$$

In isosceles triangle ACD, $$\angle CAD = \angle CDA = \alpha$$ and let $$\angle ACD = \theta$$

$$2\alpha + \theta = 180 \tag{2}$$

From two equations, we can conclude that $$\alpha = 2\beta$$. However, this does not actually help me at all. What am I missing?

Regards

@Enzo Call $$t=AC=CD=DB$$. From Pythagorean Theorem you have $$10^2=t^2+x^2$$ From the law of cosines in triangle $$ACD$$ you get $$2t^2-2t^2\cos\theta=x^2$$ But $$\cos\theta=\cos(180-2\alpha)=-\cos(2\alpha)=1-2\cos^2\alpha=1-2(t/10)^2$$. If you plug this value in the second displayed equation above you get a system for $$x$$ and $$t$$ that reduces to a quadratic equation for either $$t^2$$ or $$x^2$$
You can conclude by observing that $$\angle ADB$$ is inscribed in a semicircle and is therefore right. This means that $$\alpha + \beta = 90$$, so $$\alpha = 30$$ and $$\beta = 60$$. It follows that $$x = \frac{\sqrt{3}}{2}(10) = 5 \sqrt{3}$$.
$$\alpha + \beta + 90 = 180, \, \beta= 2 \alpha,\,3 \alpha= 90, \, \alpha=30,\, \beta=60$$
Use Rule of Sines on right triangle $$\Delta ADB$$ inside semi-circle $$ACDB$$
$$\dfrac{x}{\sin 60}= \dfrac{10}{\sin 90} \rightarrow x= 10 \sqrt{3}/2$$
• There's a typo, I meant to write $\theta = 2\beta$. How can we conclude that it is also right for $2\alpha = \beta$? – Melz Jan 22 at 18:46