Equilateral Triangle Problem: Find DP² = 3EP²

Hey guys! Let's dive into a fascinating geometry problem involving equilateral triangles and the 30°-60°-90° triangle theorem. This problem is a classic example of how geometric principles can be applied to solve mathematical puzzles. We'll break it down step by step, making it super easy to understand. So, grab your pencils and let's get started!

Understanding the Problem Statement

Before we jump into solving, let’s make sure we're all on the same page. Here’s the problem we're tackling:

We have an equilateral triangle ADEF. A line segment DPL is drawn perpendicular to the side EF. Our mission is to:

1. Draw a proper diagram representing this information.
2. Use the 30°-60°-90° triangle theorem to find the lengths of DP and EP.
3. Prove that DP² = 3EP².

Sounds like a plan? Awesome! Let’s start with the first step: drawing the diagram.

Drawing the Diagram

Alright, let's visualize what we're working with. The first step in solving any geometry problem is to draw a clear and accurate diagram. This will help us see the relationships between different parts of the figure.

1. Draw an Equilateral Triangle: Start by drawing a triangle and label its vertices as A, D, and E. Make sure all sides look equal because, remember, it’s an equilateral triangle. This means all sides (AD, DE, and EA) are of the same length, and all angles are 60°.
2. Draw the Perpendicular Line: Now, from vertex D, draw a straight line that meets side EF at a right angle (90°). This line represents DPL, where L is the point where the line intersects EF. Mark the point of intersection as P.
3. Label the Diagram: Label all the points and angles. You should have triangle ADEF with DPL perpendicular to EF. This means angle DPE and angle DPF are both 90°.

Having a well-drawn diagram is crucial because it gives us a visual aid to understand the problem better. Now that we have our diagram, we can move on to the next part: finding the lengths of DP and EP using the 30°-60°-90° triangle theorem.

Applying the 30°-60°-90° Triangle Theorem

Now, let's bring in the 30°-60°-90° triangle theorem. This theorem is a gem when dealing with right-angled triangles that have angles of 30°, 60°, and 90°. It provides a specific relationship between the sides of such triangles.

Understanding the Theorem

The 30°-60°-90° triangle theorem states that the sides are in a specific ratio:

• The side opposite the 30° angle (the shortest side) is ${ x }$.
• The side opposite the 60° angle is ${ x√3 }$.
• The side opposite the 90° angle (the hypotenuse) is ${ 2x }$.

Applying It to Our Problem

In our diagram, triangle DPE is a right-angled triangle because DP is perpendicular to EF. Since triangle ADEF is equilateral, we know that angle DEF is 60°. When we draw DPL perpendicular to EF, it bisects angle EDF, creating a 30° angle (angle EDP). So, we have a 30°-60°-90° triangle DPE.

Let’s assume the side DE of the equilateral triangle is ${ 2x }$. This makes our calculations easier.

1. Finding EP: EP is opposite the 30° angle (angle EDP). According to the theorem, this side is ${ x }$.
2. Finding DP: DP is opposite the 60° angle (angle DEF). According to the theorem, this side is ${ x√3 }$.

So, we’ve found that EP = ${ x }$ and DP = ${ x√3 }$. Great job, guys! We're one step closer to solving the problem. Now, let’s move on to the final part: proving that DP² = 3EP².

Proving DP² = 3EP²

Okay, this is the final showdown! We need to prove that DP² = 3EP². We already found the lengths of DP and EP using the 30°-60°-90° triangle theorem. Now, we just need to use those values to prove the equation.

Recall Our Findings

From the previous step, we know:

• DP = ${ x√3 }$
• EP = ${ x }$

Squaring DP and EP

Let’s start by squaring both DP and EP:

• DP² = ${ (x√3)² = x² * (√3)² = 3x² }$
• EP² = ${ x² }$

Proving the Equation

Now, we want to show that DP² = 3EP². We have DP² = ${ 3x² }$ and EP² = ${ x² }$. Let’s multiply EP² by 3:

• 3EP² = ${ 3 * x² = 3x² }$

Conclusion

Look at that! We found that DP² = ${ 3x² }$ and 3EP² = ${ 3x² }$. Therefore, DP² is indeed equal to 3EP².

We did it! We’ve successfully proven that DP² = 3EP². This is a classic example of how geometric principles and theorems can be used to solve problems.

Final Thoughts and Tips

Geometry can seem daunting, but breaking down problems into smaller steps makes them much easier to handle. Here are a few tips to keep in mind when tackling geometry problems:

• Draw Accurate Diagrams: A clear diagram is half the battle. It helps you visualize the problem and identify relationships between different parts.
• Remember Key Theorems: Theorems like the 30°-60°-90° triangle theorem are powerful tools. Make sure you understand them and know when to apply them.
• Break It Down: Complex problems can be simplified by breaking them down into smaller, manageable steps.
• Practice Makes Perfect: The more you practice, the more comfortable you’ll become with geometric concepts and problem-solving techniques.

Keep practicing, and you’ll become a geometry pro in no time! This equilateral triangle problem is just one example of the many fascinating challenges that geometry offers. Keep exploring, keep learning, and most importantly, have fun with it! You've got this!