In the quadrilateral shown below, I have two known angles ($\angle ABC = 135^\circ$ and $\angle DAB = 90^\circ$) and two known sides ($AB$ and $AD$). The side lengths have the following constraint: $AB + AD = BC + CD$.
I am trying find equations for the side lengths $BC$ and $CD$ based on $AB$ and $AD$.
I have tried working through the triangles using sine and cosine method, but unfortunately it has been a long time since I used my high school trig and algebra and I haven't figured it out.
Any help would be much appreciated.
Thanks!
This first solution doesn't work out. Skip past the horizontal divider for one that works, or if you can see why this breaks, please post a comment!
As mentioned in the hint in the comments, drop a perpendicular to $AD$ from $C$ and call the resulting intersection $H$. We may then draw in a line parallel to $AD$ passing through $B$ which intersects $CH$ at $G$. We then observe the following relations:
First, as $H$ is inside the segment $AD$, we have $DH = AD - AH$.
Next, by examining the 45-45-90 triangle $BCG$ and noting that $BG=AH$, we get $AH=BC/\sqrt{2}$ and $CH = AB + BC/\sqrt{2}$.
Then, as $CDH$ is a right triangle, we have $CH^2+DH^2=CD^2$.
Finally, we recall the equation $BC+CD=AB+AD$ from the problem statement and rewrite it as $CD = AB+AD-BC$.
Now, substituting in to equation relating $CH$, $DH$, and $CD$ we get
$$(AB+BC/\sqrt{2})^2 + (AD-BC/\sqrt{2})^2= (AB+AD-BC)^2$$
which after some light algebra gives us that
$$ BC = \frac{2(AB)(AD)}{(2+\sqrt{2})(AB+AD)} $$ and then we're finished as we already know both $AB$ and $AD$.
Somehow the above solution produces something incorrect in an example. Let's try again to see where we made a mistake.
The condition $AB+AD=BC+CD$ with $A,B,D$ fixed as in the problem determines an ellipse of possible locations for $C$. Namely, if we set $A=(0,0)$, $B=(0,b)$, and $D=(d,0)$ then the locus of $C=(x,y)$ satisfies the equation $\sqrt{x^2+(y-b)^2}+\sqrt{(x-d)^2+y^2} = b+d$, or equivalently in standard form for an ellipse, $$ (b^2+2bd)x^2+(2bd)xy+(2bd+d^2)y^2+(-2b^2d-2bd^2)x+(-2b^2d-2bd^2)y=0$$
Enforcing that the angle $ABC$ is $135$ means that $y=x+b$. Substituting this in to the equation, we see that $(b^2+6bd+d^2)x^2+(2b^2d-2bd^2)x-b^2d^2=0$, so $x=\frac{(1\pm\sqrt{2})bd^2+(-1\pm\sqrt{2})b^2d}{b^2+6bd+d^2}$, and $BC=\sqrt{2}x=\frac{(\sqrt{2}\pm2)bd^2+(-\sqrt{2}\pm2)b^2d}{b^2+6bd+d^2}$ and we can backsolve from here.
Applying this to the example of $b=100,d=40$ from the comment does indeed produce $C\approx(15.5,115.5)$ which gives $BC\approx21.93$, $CD\approx 118.07$ and both ways of calcuating $CD$ give the same result.
My best guess for an explanation as to what's gone wrong here is that somehow when putting the equations together, the $(BC)^2$ terms in the first solution canceled when they shouldn't have. I can't remember the name for why this happens - I know I've seen it once or twice before, but not in a while. If any reader can shed some light on this, please drop a line in the comments.