How to solve without using trig?

Question

In the diagram, five identical squares have been placed together.

What is $\angle ABC$?

It's easy with trig but can't find an answer without using it. Thanks!

Answer 1

Let $BD$ be an altitude of $\Delta ABC$, $E$ be a mid-point of $DC$, $\Delta AEF\sim\Delta CDB$,

such that $B$ and $F$ are placed at the different sides respect to $AC$, and $BG$ be an altitude of $\Delta BFE$.

Thus, since $EF=2BD$ and $DBGE$ is a square, we obtain: $$\Delta ABD\cong\Delta FBG,$$ which gives $$AB=BF,$$ $$\measuredangle ABF=90^{\circ}$$ and $$\measuredangle BAF=45^{\circ}.$$ Id est, $$\measuredangle ABC=180^{\circ}-\measuredangle BAC-\measuredangle BCA=180^{\circ}-\measuredangle BAC-\measuredangle FAC=180^{\circ}-\measuredangle BAF=135^{\circ}.$$

Answer 2

Here's the solution I hinted-at in the comments: