# How to solve without using trig?

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!

• Hint: Reflect point $C$ about point $B$. – Blue Jun 7 '20 at 9:10
• I've done that, but not quite sure how to proceed – Drwhops Jun 7 '20 at 9:17
• If $D$ is the reflection of $C$ about $B$, compare $|AD|$ and $|BD$|. – Blue Jun 7 '20 at 9:19
• Ah so is ADB 90 degrees, and since it is isosceles the other angles are 45 respectively? – Drwhops Jun 7 '20 at 9:20
• Glad to help. Write your own solution as an answer so that the question doesn't linger in the unanswered queue. – Blue Jun 7 '20 at 9:26

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, $$\measuredangle ABC=180^{\circ}-\measuredangle BAC-\measuredangle BCA=180^{\circ}-\measuredangle BAC-\measuredangle FAC=180^{\circ}-\measuredangle BAF=135^{\circ}.$$