No, I could not. Most angles can be found through basic Geometry, but the last two (including the desired $\angle BDA$) will require Trigonometry. I did try a system of equations based on $(1)$ and $(4)$, but it got to $68^\circ=B\hat{D}A+68^\circ-B\hat{D}A\iff0=0$: nothing useful.
Let's name the point of the crossing diagonals $O$.
$$
C\hat{O}B=180^\circ-B\hat{C}A-D\hat{B}C=112^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
B\hat{O}A=180^\circ-C\hat{O}B=68^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\
D\hat{O}C=B\hat{O}A=68^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\\
A\hat{O}D=C\hat{O}B=112^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)\\
C\hat{D}B=180^\circ-A\hat{C}D-D\hat{O}C=64^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
B\hat{A}C=180^\circ-B\hat{O}A-A\hat{B}D=74^\circ\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\
$$
(the numbers between parentheses are the basic Geometry rules used - referred below. I use the hat to refer to the vertice of the angle, and the 3-letter combination to refer to the amplitude of the respective angle, being the letters' order always counterclockwise)
$(1)$ the sum of the internal angles of any triangle is $180^\circ$
$(2)$ supplementary angles sum to $180^\circ$
$(3)$ vertically opposite angles are equal in amplitude
$(4)$ the sum of the internal angles of any quadrilateral is $360^\circ$
By $(1)$ (and $(4)$ works too) we know $68^\circ=D\hat{A}C+B\hat{D}A$, however, neither algebra nor basic Geometry take us any further (I thought there was a rule in Geometry for this, however, it requires the existence of some parallel lines, like the ones in a trapezoid).
The way harder method you used requires Trigonometry, so you certainly know all the rest (I can add it, but you didn't request it, and I'm a little rusty in that area).
This is a complete proof, and it might get a bit repetitive because I'm using the Sine Law again and again. I'm trying to find something more elegant, but this is the only thing thing I could come up with right now.( I've used a different diagram because my internet wasn't working while I was working this out. Just skip over to the ending note if you just want to see how the angles are equal.)
$DEGF$ is the quadrilateral and $A$ and $B$ are the foci.
Using sine law in $\Delta DBE$ you get $\dfrac{DE}{BE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}$ and in $\Delta GBE$ you get $\dfrac{GE}{BE}=\dfrac{\sin\angle GBE}{\sin\angle BGE}$. Dividing the two you'll obtain: $\dfrac{DE}{GE}=\dfrac{\sin\angle DBE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BGE}{\sin\angle GBE}\tag{i}$
Using sine law in $\Delta BDF$ you get $\dfrac{FD}{FB}=\dfrac{\sin\angle DBF}{\sin\angle BDF}$ and in $\Delta BGF$ you get $\dfrac{FG}{FB}=\dfrac{\sin\angle GBF}{\sin\angle BGF}$. Dividing the two you'll obtain: $\dfrac{FD}{FG}=\dfrac{\sin\angle DBF}{\sin\angle BDF}\cdot\dfrac{\sin\angle BGF}{\sin\angle GBF}\tag{ii}$
Dividing (i) and (ii):$\dfrac{DE\cdot FG}{GE\cdot FD}=\dfrac{\sin\angle BGE}{\sin\angle BDE}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}\tag{iii}$
Using sine law in $\Delta DFI$ you get $\dfrac{DI}{FI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}$ and in $\Delta GFI$ you get $\dfrac{GI}{FI}=\dfrac{\sin\angle GFE}{\sin\angle FGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}\tag{iv}$
Using sine law in $\Delta DEI$ you get $\dfrac{DI}{EI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}$ and in $\Delta GEI$ you get $\dfrac{GI}{EI}=\dfrac{\sin\angle GEF}{\sin\angle EGD}$. Dividing the two you'll obtain: $\dfrac{DI}{GI}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{v}$
Combining (iv) and (v): $\dfrac{DI}{GI}=\dfrac{\sin\angle DFE}{\sin\angle FDG}\cdot\dfrac{\sin\angle FGD}{\sin\angle GFE}=\dfrac{\sin\angle DEF}{\sin\angle EDG}\cdot\dfrac{\sin\angle EGD}{\sin\angle GEF}\tag{vi}$
Doing the same thing with $DB$ and $GB$ you will get $$\dfrac{DB}{GB}=\dfrac{\sin\angle DFE}{\sin\angle GFE}\dfrac{\sin\angle BGF}{\sin\angle BDF}=\dfrac{\sin\angle DEF}{\sin\angle GEF}\dfrac{\sin\angle BGE}{\sin\angle BDE}\tag{vii}$$
Dividing (vi) and (vii) you get:$$\dfrac{\sin\angle FGD}{\sin\angle FDG}\cdot\dfrac{\sin\angle BDF}{\sin\angle BGF}=\dfrac{\sin\angle DGE}{\sin\angle GDE}\cdot\dfrac{\sin\angle BDE}{\sin\angle BGE}\tag{viii}$$
Now due to the Property 1 on this link, $\angle FGD=\angle BGE, \angle GDE=\angle BDF,\angle DGE=\angle BGF$ and $\angle BDE=\angle FDG$. Making the necessary substitutions, relation (viii) becomes:$$\sin^2\angle BGE \cdot\sin^2\angle BDF=\sin^2\angle BDE \cdot\sin^2\angle BGF$$ $$\sin\angle BGE \cdot\sin\angle BDF=\sin\angle BDE \cdot\sin\angle BGF\tag{ix}$$
Combining (iii) and (ix): $DE\cdot FG=GE\cdot FD$
NOTE: Making use of the relations (vi),(vii) and (ix) it is easy to see that $\dfrac{DI}{GI}=\dfrac{DB}{GB}$. By the Angle Bisector Theorem, this means that $\angle DBF = \angle GBF$ and similarly $\angle FAG = \angle EAG$.
Best Answer
The three perpendiculars from the corners of a triangle to the sides meet in a common point, the orthocentre. You know that $UP$ is perpendicular to $RT$ and $TQ$ is perpendicular to $RU$; it follows that the line through $R$ and their intersection $S$ is perpendicular to $TU$.
[Update:] It turns out you can actually go on deducing all the angles much like you started. I'll use your angles $1$, $2$ and $3$. (I don't understand how you defined $4$; in the drawing it seems to mark the right angles $\angle SPT$ and $\angle SQU$ but in the text you defined it as $\angle PQS$.)
First, to make the symmetry of the situation manifest, let's also draw the lines $PA'$ and $QA'$. Here's an image (I'll be justifying the angles I filled in in a bit):
The triangle $PQA'$ is the orthic triangle of the triangle $RTU$. If you take out the circle $\Gamma$, the diagram has $S_3$ symmetry, so the quadrilaterals $A'URP$ and $QRTA'$ are cyclic for the same reason as $PTUQ$; that justfies the angles I've filled in. Since $|OR|=|OP|=r$, triangle $ROP$ is isosceles, so $\angle OPR$ is 3, and hence $\angle OPA$ is 1. Thus the triangles $OPA$ and $OPA'$ have two angles in common (1 at $\angle OPA$ and $\angle OA'P$, and $\angle POA=\angle POA'$), and hence are similar. The inversion property then follows by taking the ratios of corresponding sides in these triangles.
P.S.: That the altitudes of $RTU$ are the bisectors of its orthic triangle is related to the fact (mentioned in the Wikipedia article) that the orthocentre is the incentre of the orthic triangle.