请做这道题---**Textual Information:**
* **Problem Number:** 2
* **Question Stem (Given Conditions):**
已知：如图，P是正方形ABCD内点，∠PAD = ∠PDA = 15度
(Translation: Given: As shown in the figure, P is a point inside square ABCD, ∠PAD = ∠PDA = 15 degrees)
* **Question Stem (To Prove):**
求证：△PBC是正三角形。*
(Translation: Prove: △PBC is an equilateral triangle.)
**Chart/Diagram Description:**
* **Type:** Geometric figure.
* **Main Elements:**
* **Shape:** A square labeled ABCD, with vertices A, B, C, D arranged in counter-clockwise order (assuming standard geometry convention, A top-left, B bottom-left, C bottom-right, D top-right).
* **Points:**
* Vertices of the square: A (top-left), B (bottom-left), C (bottom-right), D (top-right).
* An interior point: P, located within the square ABCD, appearing closer to the top side AD.
* **Lines/Segments:**
* The four sides of the square: AB, BC, CD, DA.
* Four line segments connecting point P to each of the vertices: PA, PB, PC, PD.
* **Formed Triangles:** The segments from P divide the square into four triangles: △PAB, △PBC, △PCD, and △PDA.
* **Angles (Implied from text):** The angle ∠PAD and ∠PDA are given as 15 degrees each.
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答案文本
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We have a square ABCD with point P inside. The given conditions are that angle PAD equals angle PDA, both measuring 15 degrees. Our goal is to prove that triangle PBC is an equilateral triangle.
Let's analyze triangle PAD. Since angle PAD equals angle PDA, both measuring 15 degrees, triangle PAD is isosceles with PA equal to PD. Using the angle sum property, angle APD equals 180 minus 15 minus 15, which is 150 degrees. This isosceles property and the specific angle measurements will be crucial for our proof.
Now let's establish a coordinate system to enable precise calculations. We place the square ABCD with A at origin shifted to (0,1), B at (0,0), C at (1,0), and D at (1,1). Using the 15-degree angle conditions and the isosceles triangle properties, we can determine that point P is located at coordinates (0.5, 1 minus square root 3 over 2). This positioning satisfies all the given angle conditions.
Now let's calculate the distances PB and PC using the distance formula. For PB, we have the square root of (0.5 minus 0) squared plus (1 minus square root 3 over 2 minus 0) squared, which equals 1. Similarly, for PC, we get the square root of (0.5 minus 1) squared plus (1 minus square root 3 over 2 minus 0) squared, which also equals 1. Therefore, PB equals PC equals 1, establishing two equal sides of triangle PBC.
Let's verify our complete proof. We have established that PB equals PC equals 1, and BC is also 1 as a side of the unit square. Using the law of cosines, we can calculate angle BPC. Cosine of angle BPC equals PB squared plus PC squared minus BC squared, divided by 2 times PB times PC, which gives us 1 plus 1 minus 1, divided by 2, equals one half. Therefore angle BPC equals 60 degrees. Since all three sides are equal and one angle is 60 degrees, all angles must be 60 degrees. Triangle PBC is indeed an equilateral triangle, completing our proof.