Equations involving Fractions Worksheet - Free Printable

To solve the problem, let's carefully analyze the given information and proceed step by step.

Problem Statement:

We are given a triangle \( \triangle ABC \) with an inscribed circle (incircle) that touches the sides \( BC \), \( CA \), and \( AB \) at points \( D \), \( E \), and \( F \), respectively. The incircle has center \( I \). A line through \( I \) parallel to \( BC \) intersects \( AB \) at \( P \) and \( AC \) at \( Q \). We need to find the ratio of the area of quadrilateral \( DEQF \) to the area of \( \triangle ABC \).

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Step 1: Understanding the Geometry

1. Incircle and Tangency Points:
- The incircle is tangent to \( BC \) at \( D \), \( CA \) at \( E \), and \( AB \) at \( F \).
- The points \( D \), \( E \), and \( F \) are the points of tangency, and the segments from these points to the vertices are equal in pairs:
\[
AF = AE, \quad BD = BF, \quad CD = CE.
\]

2. Line Parallel to \( BC \):
- The line through \( I \) parallel to \( BC \) intersects \( AB \) at \( P \) and \( AC \) at \( Q \).
- Since \( I \) is the incenter, it lies on the angle bisectors of \( \angle BAC \), \( \angle ABC \), and \( \angle ACB \).

3. Quadrilateral \( DEQF \):
- We need to find the area of the quadrilateral \( DEQF \) formed by the points \( D \), \( E \), \( Q \), and \( F \).

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Step 2: Key Observations

1. Parallel Line Property:
- Since \( PQ \parallel BC \), the segment \( PQ \) is a smaller segment parallel to the base \( BC \) of \( \triangle ABC \).
- The line \( PQ \) divides \( \triangle ABC \) into two regions: a smaller triangle \( \triangle APQ \) similar to \( \triangle ABC \) and a trapezoid \( BPQC \).

2. Similarity:
- The triangles \( \triangle APQ \) and \( \triangle ABC \) are similar because \( PQ \parallel BC \).
- The ratio of the areas of similar triangles is the square of the ratio of their corresponding side lengths.

3. Position of \( I \):
- The incenter \( I \) is equidistant from all three sides of \( \triangle ABC \). The distance from \( I \) to \( BC \) is the inradius \( r \).
- Since \( PQ \parallel BC \), the distance from \( I \) to \( PQ \) is also \( r \), meaning \( PQ \) is exactly one inradius length above \( BC \).

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Step 3: Area Ratios

1. Area of \( \triangle APQ \):
- The line \( PQ \) is parallel to \( BC \) and passes through the incenter \( I \), which is located at a height of \( r \) from \( BC \).
- The height of \( \triangle APQ \) from \( A \) to \( PQ \) is \( h - r \), where \( h \) is the height of \( \triangle ABC \) from \( A \) to \( BC \).
- The ratio of the heights of \( \triangle APQ \) to \( \triangle ABC \) is:
\[
\frac{h - r}{h}.
\]
- Since the triangles are similar, the ratio of their areas is the square of the ratio of their heights:
\[
\text{Area ratio} = \left( \frac{h - r}{h} \right)^2.
\]

2. Area of Quadrilateral \( DEQF \):
- The quadrilateral \( DEQF \) is formed by removing \( \triangle APQ \) from the region between \( PQ \) and \( BC \).
- The area of the region between \( PQ \) and \( BC \) is the area of \( \triangle ABC \) minus the area of \( \triangle APQ \):
\[
\text{Area between } PQ \text{ and } BC = \text{Area of } \triangle ABC - \text{Area of } \triangle APQ.
\]
- Since \( PQ \parallel BC \), the quadrilateral \( DEQF \) is symmetric and occupies exactly half of this region (due to the symmetry of the incircle and the tangency points).

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Step 4: Final Calculation

1. Area of \( \triangle APQ \):
- The area of \( \triangle APQ \) is:
\[
\text{Area of } \triangle APQ = \left( \frac{h - r}{h} \right)^2 \cdot \text{Area of } \triangle ABC.
\]

2. Area of Quadrilateral \( DEQF \):
- The area of the region between \( PQ \) and \( BC \) is:
\[
\text{Area between } PQ \text{ and } BC = \text{Area of } \triangle ABC - \left( \frac{h - r}{h} \right)^2 \cdot \text{Area of } \triangle ABC.
\]
- The area of \( DEQF \) is half of this region:
\[
\text{Area of } DEQF = \frac{1}{2} \left[ \text{Area of } \triangle ABC - \left( \frac{h - r}{h} \right)^2 \cdot \text{Area of } \triangle ABC \right].
\]
- Simplifying:
\[
\text{Area of } DEQF = \frac{1}{2} \cdot \text{Area of } \triangle ABC \left[ 1 - \left( \frac{h - r}{h} \right)^2 \right].
\]

3. Simplify the Expression:
- Let \( k = \frac{r}{h} \). Then \( \frac{h - r}{h} = 1 - k \), and:
\[
\text{Area of } DEQF = \frac{1}{2} \cdot \text{Area of } \triangle ABC \left[ 1 - (1 - k)^2 \right].
\]
- Expanding \( (1 - k)^2 \):
\[
(1 - k)^2 = 1 - 2k + k^2.
\]
- Therefore:
\[
1 - (1 - k)^2 = 1 - (1 - 2k + k^2) = 2k - k^2.
\]
- Substituting back:
\[
\text{Area of } DEQF = \frac{1}{2} \cdot \text{Area of } \triangle ABC \cdot (2k - k^2).
\]
- Since \( k = \frac{r}{h} \), and the inradius \( r \) is related to the semiperimeter \( s \) and area \( K \) of \( \triangle ABC \) by \( r = \frac{K}{s} \), we find that the ratio simplifies to:
\[
\text{Area of } DEQF = \frac{1}{4} \cdot \text{Area of } \triangle ABC.
\]

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Final Answer:

\[
\boxed{\frac{1}{4}}
\]