Finding the Relationship Between $∠GCF$ and $α$ in a Rhombus with an Isosceles Triangle

Question

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.

Answer 1

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 $q:=|BE|$.

• 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, $$\angle DCF\;=\;\angle HCD-\angle HCF\;=\;2\beta-\gamma\;=\;\frac32\angle ABC-90^\circ$$

Answer 2

If you look properly, $$\angle AFE=\angle FAE=\frac{180°-\angle FEA}{2}=\frac{180°-\alpha}{2}$$

Thus for $\triangle GCF$, $$\angle GCF+\angle GFC+\angle CGF=180°$$ OR $$\angle GCF + \frac{180°-\alpha}{2}+\angle EFC+\angle CGF= 180°$$

Now $\angle BAG=\angle BAE+\angle EAF=\angle CGF$(corresponding angles)

So finally $$\angle GCF + \frac{180°-\alpha}{2}+\angle BAE+\angle EFC+ \angle EAF= 180°$$ $$\implies \angle GCF+180°-\alpha+\angle BAE+\angle EFC=180°$$ $$\implies \angle GCF +\angle BAE+ \angle EFC= \alpha$$

Answer 3

Let $a=\angle BAE=\angle CAF$. As $|AB|=k|AC|$, we have

$$a|AB|=ak|AC|$$

and by the sine rule, therefore

$$\angle AEB |AE|=k\angle CFA|AF|$$

Then we have $\triangle ABE\sim\triangle ACF$ by the power series expansion of the loci and angles.

Answer 4

One effective method to address the problem involves a geometric construction:

1. Constructing Line AC: Start by connecting point A to point C in the rhombus. This line will play a crucial role in our subsequent geometric analysis.

2. Identifying Similar Triangles: Observe that triangle ABE is similar to triangle ACF. This similarity is crucial as it implies that corresponding angles in these triangles are equal.

3. Equating Angles: Due to the similarity between triangles ABE and ACF, we can conclude that angle ACF is equal to angle ABC. Given that angle ABC is given as α, angle ACF must also be α.

4. Calculating Angle GCF: To find angle GCF, consider the angle difference at point C. Angle GCF is calculated as angle ACF minus angle ACD. Given that angle ACD (the angle between line AC and line CD) is half of the angle opposite to it in the rhombus, which is ($90^\circ - \frac{1}{2}\alpha$), the calculation follows:

$\text{Angle GCF} = \alpha - \left(90^\circ - \frac{1}{2}\alpha\right) = \frac{3}{2}\alpha - 90^\circ.$