What is the value for angle EGB?

Question

(OBM) In a rectangle ABCD, E is the midpoint of side BC and F is the midpoint of side CD. G is the point that represents the intersection between DE and FB. Angle EAF is 20º. What is the value for the angle EGB? Sorry, I can't photograph my attempts…

Answer 1

Regardless what the angle of $\angle EAF$ is, $\angle EGB = \angle EAF$ is always true. Observe that $\angle IDF = \angle BAH$ and $\angle ABH = 90^\circ - \angle CBF = 90^\circ - \angle DAF = \angle DFI $. This gives us $\triangle ABH \sim \triangle DFI$, and $$\angle AHB = \angle DIF = \angle AIG \implies AIGH \ \text{is cyclic} \implies \angle EGB = \angle EAF$$ Hence, in this case, $\angle EGB = \angle EAF = 20^\circ$.

Answer 2

Let $\angle BAE=\alpha$ and $\angle DAF=\beta$. Then $\alpha+\beta=90^\circ-20^\circ=70^\circ$.

Note that $\triangle ABE\cong\triangle DCE$ and $\triangle ADF\cong \triangle BCF$.

So $\angle CDE=\angle BAE=\alpha$ and $\angle CBF=\angle DAF=\beta$.

$\angle BEG=\angle BAE+\angle BCD=\alpha+90^\circ$.

$\angle EGB=180^\circ-\angle CBF-\angle BEG=180^\circ-(\alpha+90^\circ)-\beta=90^\circ-\alpha-\beta=20^\circ$.