# Through the midpoint of the diagonal ac of the rectangle abcd, a straight line is drawn that intersects the sides BC

**Through the midpoint of the diagonal ac of the rectangle abcd, a straight line is drawn that intersects the sides BC and AD at points P and K, respectively. 1) Prove that APCK is a parallelogram. 2) Find the area APCK if AK = 4, KD = 8 and AC = 13. 3) Find PK. 4) Use a micro calculator to find the AOK angle.**

Exercise 1.

Let a rectangle ABCD be given, its diagonal AC is divided by point O into two equal segments AO = OC = AC / 2. A straight line is drawn through tO, which intersects BC at t P and AD at point K, let us prove that AРСK is a parallelogram.

A sign that a quadrilateral is a parallelogram is that its intersecting diagonals divide each other in half. Our AС is divided in half, we will prove that the RK is also divided in half, RK / 2 = PO = OK.

1. Consider triangles AOK and POC, AO = OC, <OAK = <PCO, as internal versatile at the intersection of two straight lines – a secant, <AOK = <POC, as vertical angles at the intersection of two straight lines.

Triangles AOK and POC are equal, according to the second criterion of equality of triangles: If the side and the angles adjacent to it of one triangle are equal, respectively, to the side and the angles adjacent to it of the other triangle, then such triangles are equal. Means: AK = RS, OK = RO.

2. We have AO = OС, OK = RO, which means both diagonals are divided in half, which means this is a parallelogram.

Task 2.

Let’s find the area of the parallelogram ARSK. By the property of a parallelogram, its diagonal divides into two equal triangles, AKC = APC, which means S APCK = 2 * S AKC.

Consider a triangle AKC, AK = 4 cm, AC = 13 cm, find the side of the COP.

Consider a triangle ACD it is rectangular, <D = 90 °, AD = AK + KD = 4 + 8 = 12 cm AC = 13 cm.By the Pythagorean theorem:

CD = √ (AC²-AD²) = √ (13²-12²) = √ (169-144) = √25 = 5 cm.

Consider a triangle CKD, it is rectangular, <D = 90 °, KD = 8 cm, CD = 5 cm.By the Pythagorean theorem:

KS = √ (KD² + CD²) = √ (8² + 5²) = √ (64 + 25) = √89 = 9.44 cm.

Let’s return to the triangle AKС, AK = 4 cm, AC = 13 cm, KS = 9.44 cm, according to Heron’s theorem:

S = √ (p (p-AK) (p-AС) (p-KС)), we find the semiperimeter:

p = (AK + AC + KС) / 2 = (4 + 13 + 9.44) / 2 = 13.22 cm.

Find the area:

S = √ (p (p-AK) (p-AС) (p-KС)) = √ (13.22 * (13.22-4) (13.22-13) (13.22-9.44 )) = 10 cm².

Let’s find the area of the parallelogram AРСK:

S ARSK = 2 * S AKS = 2 * 10 = 20 cm².

Answer: the area of the parallelogram ARSK is 20 cm².

Task 3.

By the property of the diagonals of a parallelogram, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides:

PK² + AC² = 2AK² + 2CK².

PK = √ (2AK² + 2CK²-AC²) = √ (2 * 4² + 2 * (√89) ²-13²) = √41 cm.

Answer: РK = √41 cm or 6.4 cm.

Task 4.

The parallelogram area is determined by the formula:

S = (d1 * d1 * sin ɣ) / 2 = (AC * PK * sin AOK) / 2.

Let us express the sine of the angle from this expression:

sin AOK = 2 * S / (AC * PK) = 2 * 20 / (13 * 6.4) = 0.48.

Take the arcsine to determine the degree measure of the angle:

arcsin (0.48) = 29 °.

Answer: <AOK = 29 °.