For triangle ABC, DC = DE = DH, Find the value of angle X.

Question

In the triangle ABC, DC = DE = DH, Find the value of angle X.

I found the right triangles AHC and BHC, I also found the angle ACH = 40°. I found the isosceles triangles HDE, DEC, and DHC, which means the angle DHE = HED, the angle DEC = DCE, and the angle DHC = DCH.

Thanks in advance.

Answer 1

Let $\measuredangle DCH = z$. Then $\lvert DH\rvert = \lvert DC\rvert$ means

$$\measuredangle DHC = z$$

Also, since as you've already noted, $\measuredangle ACH = 40^{\circ}$, we have that since $\lvert DC\rvert = \lvert DE\rvert$, then

$$\measuredangle DEC = 40^{\circ} + z \;\;\to\;\; \measuredangle HED = 140^{\circ} - z$$

Using that the sum of angles in any quadrilateral is $360^{\circ}$, plus the above $2$ results, we then have from $AEDH$ that

$$X = 360^{\circ} - (90^{\circ} + z) - 50^{\circ} - (140^{\circ} - z) = 80^{\circ}$$

Your second diagram's estimate of about $77.579^{\circ}$ is reasonably close.

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Somewhat more directly, as Intelligenti pauca's comment indicates, since it's given that $\lvert DC\rvert = \lvert DE\rvert = \lvert DH\rvert = r$, we can draw a circle through the points $C$, $E$ and $H$, where $D$ is the center and $r$ is its radius. The central angle theorem states that

The central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.

In your case, using $HE$ as the base, $X$ is the central angle while an inscribed angle is $\angle ECH = 40^{\circ}$, as you've already noted, so then

$$X = 2(40^{\circ}) = 80^{\circ}$$