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

enter image description here

What is $\angle ABC$?

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

  • $\begingroup$ Hint: Reflect point $C$ about point $B$. $\endgroup$ – Blue Jun 7 '20 at 9:10
  • $\begingroup$ I've done that, but not quite sure how to proceed $\endgroup$ – Drwhops Jun 7 '20 at 9:17
  • $\begingroup$ If $D$ is the reflection of $C$ about $B$, compare $|AD|$ and $|BD$|. $\endgroup$ – Blue Jun 7 '20 at 9:19
  • 3
    $\begingroup$ Ah so is ADB 90 degrees, and since it is isosceles the other angles are 45 respectively? $\endgroup$ – Drwhops Jun 7 '20 at 9:20
  • 1
    $\begingroup$ Glad to help. Write your own solution as an answer so that the question doesn't linger in the unanswered queue. $\endgroup$ – Blue Jun 7 '20 at 9:26

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

enter image description here


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


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