Geometry – Relationship Between GCF and Rhombus with Isosceles Triangle

euclidean-geometrygeometrytriangles

I am working on a problem involving a rhombus $ABCD$, where point $E$ lies on side $BC$. Triangle $AEF$ is an isosceles triangle with $AE = EF$, and $∠AEF = ∠ABC = α$, where $α$ is at least $90°$. Line $AF$ intersects line $CD$ at point $G$. I need to find the relationship between angle $∠GCF$ and $α$.

Rhombus Properties: In a rhombus, all sides are equal and opposite angles are equal. The diagonals intersect at right angles and bisect the vertex angles.

Given that $α ≥ 90°$, which indicates that angle $α$ is at least a right angle, I am particularly interested in how this impacts the angle $∠GCF$ at the intersection.

How does angle $∠GCF$ relate to $α$ in this geometric configuration? Any insights or geometric proofs are greatly appreciated.
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Best Answer

Changing some notation, I let $\angle ABC=2\beta$ and $\angle DCB=2\gamma$; note that $\beta+\gamma=90^\circ$. Also, define $p:=|AB|$ and $q:=|BE|$.

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  • Extend $\overline{BC}$ by length $q$ to a point $H$, so that $|AB|=|EH|=p$. Since $\angle BAE\cong\angle CEF$ (why?), we have $\triangle ABE\cong\triangle EHF$ by SAS.

  • We also have $|CH|=|FH|=q$, making $\triangle CHF$ isosceles.

  • We conclude that $$\angle HCF\;=\;\gamma\;=\;\frac12\angle BCD$$ or, with a bit of angle arithmetic, $$\angle DCF\;=\;\angle HCD-\angle HCF\;=\;2\beta-\gamma\;=\;\frac32\angle ABC-90^\circ$$

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