Are these triangles similar?

Question

Given only the information conveyed in the diagram (cannot assume any angle measure except the 2 right angles explicitly specified), are triangles AED and DFC similar? My thinking is they are not guaranteed to be similar because angles EBF and EDF are not necessarily 90 degrees.

Answer 1

The six points highlighted in the photo can be parameterized as follows:

$$ A = (0,0), \quad \quad \quad B = \left(b\cos\alpha,\,b\sin\alpha\right), \quad \quad \quad C = \left(c,\,0\right), $$

$$ D = A + (C-A)\,t, \quad \quad \quad E = A + (B-A)\,u, \quad \quad \quad F = B + (C-B)\,v, $$

with $0 < \alpha < \frac{\pi}{2}$, $b>0$, $c>0$, $0<t<1$, $0<u<1$, $0<v<1$.

So, imposing the two orthogonality conditions:

$$ \begin{aligned} & (E-A)\cdot(E-D) = 0 \quad \quad \Leftrightarrow \quad \quad u = \frac{c\,t\cos\alpha}{b}\,; \\ & (F-C)\cdot(F-D) = 0 \quad \quad \Leftrightarrow \quad \quad v = \frac{b(b-c\cos\alpha)+c\,t(c-b\cos\alpha)}{b^2+c^2-2\,b\,c\cos\alpha}\,; \\ \end{aligned} $$

it's possible to impose that the corresponding sides are in proportion:

$$ \frac{\overline{AD}}{\overline{DC}} = \frac{\overline{AE}}{\overline{DF}} = \frac{\overline{DE}}{\overline{CF}} $$

which together with the conditions of existence of the six parameters leads to:

$$ b = c\cos\alpha, \quad \quad \text{with} \; 0 < \alpha < \frac{\pi}{2}, \; c > 0, \; 0 < t < 1\,. $$

So yes, you're right, only with this other condition can we be sure of the similarity.

Answer 2

As given there is no similarity as per a construction with simple angle chasing around a cyclic quadrilateral.

This is because no angle coincidence in the right triangles.

This can happen only if the inscribed cyclic quadrilateral is a rectangle.