Question

Очень срочно! В квадрате ABCD на диагонали AC отметили точку P так, что AP:PC = 3:1. L - середина AB. Докажите, что угол LPD равен 90 градусам.

Answer 1

Решим координатным методом. Пусть сторона квадрата равна 4. Оси координат направим по сторонам AD и AB. 
A(0,0); B(0,4); C(4,4), D(4,0), L(0,2); P(3,3).

Вектор PL(0-3,2-3)=(-3,-1); вектор PD(4-3,0-3)=(1,-3);
скалярное произведение PL·PD =(-3)·1+(-1)·(-3)=-3+3=0⇒ векторы перпендикулярны, что и требовалось доказать.

Answer 2

Proof of Angle LPD Being 90 Degrees

To prove that angle LPD is 90 degrees, we can use the properties of triangles and the given information about the quadrilateral ABCD.

Given: - Quadrilateral ABCD with diagonal AC and point P such that AP:PC = 3:1. - L is the midpoint of AB.

Proof:

1. Constructing the Diagram: - Let's start by constructing the diagram of quadrilateral ABCD with diagonal AC and point P such that AP:PC = 3:1. We'll also mark L as the midpoint of AB.

2. Using Midpoint Theorem: - The Midpoint Theorem states that a line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of its length. - In this case, since L is the midpoint of AB, it follows that LP is parallel to DC and LP = 1/2 * DC.

3. Using Similar Triangles: - Consider triangles DPC and LAP. - By the Angle Bisector Theorem, we know that AP:PC = sin ∠DAP:sin ∠DPC. - Since AP:PC = 3:1, sin ∠DAP:sin ∠DPC = 3:1. - Therefore, triangles DPC and LAP are similar by the Angle-Angle (AA) criterion.

4. Concluding the Proof: - Since triangles DPC and LAP are similar, it follows that angle LPD is equal to angle DPC. - Since LP is parallel to DC and LP = 1/2 * DC, and angle LPD = angle DPC, we can conclude that angle LPD is a right angle (90 degrees) by the properties of parallel lines and transversals.

Therefore, we have proved that angle LPD is 90 degrees.