We solve problems in geometry: solving quadrilaterals. Parallelogram in problems Proofs of some theorems

Task 4.

One of the diagonals of a parallelogram of length 4√6 makes an angle of 60° with the base, and the second diagonal makes an angle of 45° with the same base. Find the second diagonal.

Solution.

1. AO = 2√6.

2. Apply the sine theorem to the triangle AOD.

AO/sin D = OD/sin A.

2√6/sin 45 o = OD/sin 60 o.

OD = (2√6sin 60 o) / sin 45 o = (2√6 √3/2) / (√2/2) = 2√18/√2 = 6.

Answer: 12.

Task 5.

For a parallelogram with sides 5√2 and 7√2, the smaller angle between the diagonals is equal to the smaller angle of the parallelogram. Find the sum of the lengths of the diagonals.

Solution.

Let d 1, d 2 be the diagonals of the parallelogram, and the angle between the diagonals and the smaller angle of the parallelogram be φ.

1. Let's count two different
ways of its area.

S ABCD \u003d AB AD sin A \u003d 5√2 7√2 sin f,

S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin f.

We obtain the equality 5√2 7√2 sin f = 1/2d 1 d 2 sin f or

2 5√2 7√2 = d 1 d 2 ;

2. Using the ratio between the sides and diagonals of the parallelogram, we write the equality

(AB 2 + AD 2) 2 = AC 2 + BD 2.

((5√2) 2 + (7√2) 2) 2 = d 1 2 + d 2 2 .

d 1 2 + d 2 2 = 296.

3. Let's make a system:

(d 1 2 + d 2 2 = 296,
(d 1 + d 2 = 140.

Multiply the second equation of the system by 2 and add it to the first.

We get (d 1 + d 2) 2 = 576. Hence Id 1 + d 2 I = 24.

Since d 1, d 2 are the lengths of the diagonals of the parallelogram, then d 1 + d 2 = 24.

Answer: 24.

Task 6.

The sides of the parallelogram are 4 and 6. The acute angle between the diagonals is 45 o. Find the area of ​​the parallelogram.

Solution.

1. From the triangle AOB, using the cosine theorem, we write the relationship between the side of the parallelogram and the diagonals.

AB 2 \u003d AO 2 + VO 2 2 AO VO cos AOB.

4 2 \u003d (d 1 / 2) 2 + (d 2 / 2) 2 - 2 (d 1 / 2) (d 2 / 2) cos 45 o;

d 1 2/4 + d 2 2/4 - 2 (d 1/2) (d 2/2)√2/2 = 16.

d 1 2 + d 2 2 - d 1 d 2 √2 = 64.

2. Similarly, we write the relation for the triangle AOD.

We take into account that<АОD = 135 о и cos 135 о = -cos 45 о = -√2/2.

We get the equation d 1 2 + d 2 2 + d 1 d 2 √2 = 144.

3. We have a system
(d 1 2 + d 2 2 - d 1 d 2 √2 = 64,
(d 1 2 + d 2 2 + d 1 d 2 √2 = 144.

Subtracting the first from the second equation, we get 2d 1 d 2 √2 = 80 or

d 1 d 2 = 80/(2√2) = 20√2

4. S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin α \u003d 1/2 20√2 √2/2 \u003d 10.

Note: In this and in the previous problem, there is no need to solve the system completely, foreseeing that in this problem we need the product of diagonals to calculate the area.

Answer: 10.

Task 7.

The area of ​​the parallelogram is 96 and its sides are 8 and 15. Find the square of the smaller diagonal.

Solution.

1. S ABCD \u003d AB AD sin VAD. Let's do a substitution in the formula.

We get 96 = 8 15 sin VAD. Hence sin VAD = 4/5.

2. Find cos BAD. sin 2 VAD + cos 2 VAD = 1.

(4/5) 2 + cos 2 BAD = 1. cos 2 BAD = 9/25.

According to the condition of the problem, we find the length of the smaller diagonal. Diagonal BD will be smaller if angle BAD is acute. Then cos BAD = 3 / 5.

3. From the triangle ABD, using the cosine theorem, we find the square of the diagonal BD.

BD 2 \u003d AB 2 + AD 2 - 2 AB BD cos BAD.

ВD 2 \u003d 8 2 + 15 2 - 2 8 15 3 / 5 \u003d 145.

Answer: 145.

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Theorem 1. The area of ​​a trapezoid is equal to the product of half the sum of its bases and the height:

Theorem 2. The diagonals of a trapezoid divide it into four triangles, two of which are similar and the other two have the same area:

Theorem 3. The area of ​​a parallelogram is equal to the product of the base and the height lowered to the given base, or the product of the two sides and the sine of the angle between them:

Theorem 4. In a parallelogram, the sum of the squares of the diagonals is equal to the sum of the squares of its sides:

Theorem 5. The area of ​​an arbitrary convex quadrilateral is equal to half the product of its diagonals and the sine of the angle between them:

Theorem 6. The area of ​​a quadrilateral circumscribed about a circle is equal to the product of the semiperimeter of this quadrilateral and the radius of the given circle:

Theorem 7. A quadrilateral whose vertices are the midpoints of the sides of an arbitrary convex quadrilateral is a parallelogram whose area is equal to half the area of ​​the original quadrilateral:

Theorem 8. If the diagonals of a convex quadrilateral are mutually perpendicular, then the sums of the squares of the opposite sides of this quadrilateral are:

AB2 + CD2 = BC2 + AD2.

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Proofs of some theorems

Proof of Theorem 2. Let ABCD be a given trapezoid, AD and BC its bases, O the intersection point of diagonals AC and BD of this trapezoid. Let us prove that triangles AOB and COD have the same area. To do this, let us drop perpendiculars BP and CQ from points B and C to line AD. Then the area of ​​triangle ABD is

And the area of ​​triangle ACD is

Since BP = CQ, then S∆ABD = S∆ACD . But the area of ​​triangle AOB is the difference between the areas of triangles ABD and AOD, and the area of ​​triangle COD is the difference between the areas of triangles ACD and AOD. Therefore, the areas of triangles AOB and COD are equal, which was to be proved.

Proof of Theorem 4. Let ABCD be a parallelogram, AB = CD = a, AD = BC = b,
AC = d1 , BD = d2 , ∠BAD = α, ∠ADC = 180° – α. Let us apply the cosine theorem to the triangle ABD:

Applying now the cosine theorem to the triangle ACD, we get:

Adding term by term equalities, we get that Q.E.D.

Proof of Theorem 5. Let ABCD be an arbitrary convex quadrangle, E the intersection point of its diagonals, AE = a, BE = b,
CE = c, DE = d, ∠AEB = ∠CED = ϕ, ∠BEC =
= ∠AED = 180° – ϕ. We have:

Q.E.D.

Proof of Theorem 6. Let ABCD be an arbitrary quadrilateral circumscribed about a circle, O be the center of this circle, OK, OL, OM and ON be the perpendiculars dropped from the point O to the lines AB, BC, CD and AD, respectively. We have:

where r is the radius of the circle and p is the semi-perimeter of quadrilateral ABCD.

Proof of Theorem 7. Let ABCD be an arbitrary convex quadrilateral, K, L, M and N be the midpoints of the sides AB, BC, CD and AD, respectively. Since KL is the midline of triangle ABC, line KL is parallel to line AC and Similarly, line MN is parallel to line AC and Therefore, KLMN is a parallelogram. Consider triangle KBL. Its area is equal to a quarter of the area of ​​triangle ABC. The area of ​​triangle MDN is also equal to a quarter of the area of ​​triangle ACD. Consequently,

Likewise,

It means that

whence it follows that

Proof of Theorem 8. Let ABCD be an arbitrary convex quadrilateral whose diagonals are mutually perpendicular, let E be the intersection point of its diagonals,
AE= a, BE = b, CE = c, DE = d. Apply the Pythagorean theorem to triangles ABE and CDE:
AB2=AE2+BE2= a 2 + b2 ,
CD2 = CE2 + DE2 = c2 + d2,
Consequently,
AB2+CD2= a 2 + b2 + c2 + d2 .
Applying now the Pythagorean theorem to triangles ADE and BCE, we get:
AD2=AE2+DE2= a 2 + d2 ,
BC2 = BE2 + CE2 = b2 + c2 ,
whence it follows that
AD2+BC2= a 2 + b2 + c2 + d2 .
Hence, AB2 + CD2 = AD2 + BC2 , which was to be proved.

Problem solving

Task 1. A trapezoid is described near the circle with base angles α and β. Find the ratio of the area of ​​the trapezoid to the area of ​​the circle.

Solution. Let ABCD be a given trapezoid, AB and CD its bases, DK and CM the perpendiculars dropped from points C and D to line AB. The desired ratio does not depend on the radius of the circle. Therefore, we assume that the radius is 1. Then the area of ​​the circle is π, we find the area of ​​the trapezoid. Since triangle ADK is a right triangle,

Similarly, from a right triangle BCM we find that Since a circle can be inscribed in a given trapezoid, then the sums of opposite sides are equal:
AB + CD = AD + BC,
where do we find

So the area of ​​the trapezoid is

and the desired ratio is
Answer:

Task 2. In a convex quadrilateral ABCD, angle A is 90° and angle C does not exceed 90°. Perpendiculars BE and DF are dropped from vertices B and D to diagonal AC. It is known that AE = CF. Prove that angle C is a right angle.

Proof. Since angle A is 90°,
and angle C does not exceed 90°, then points E and F lie on diagonal AC. Without loss of generality, we may assume that AE< AF (в противном случае следует повторить все нижеследующие рассуждения с заменой точек B и D). Пусть ∠ABE = α,
∠EBC = β, ∠FDA = γ, ∠FDC = δ. It suffices for us to prove that α + β + γ + δ = π. Because

whence we obtain that which was to be proved.

Task 3. The perimeter of an isosceles trapezoid circumscribed about a circle is p. Find the radius of this circle if it is known that the acute angle at the base of the trapezoid is α.
Solution. Let ABCD be a given isosceles trapezoid with bases AD and BC, let BH be the height of this trapezoid from vertex B.
Since a circle can be inscribed in a given trapezoid, then

Consequently,

From right triangle ABH we find,

Answer:

Task 4. Given a trapezoid ABCD with bases AD and BC. Diagonals AC and BD intersect at point O, and lines AB and CD intersect at point K. Line KO intersects sides BC and AD at points M and N, respectively, and angle BAD is 30°. It is known that a circle can be inscribed in the trapezoids ABMN and NMCD. Find the area ratio of triangle BKC and trapezoid ABCD.

Solution. As you know, for an arbitrary trapezoid, the line connecting the intersection point of the diagonals and the intersection point of the extensions of the lateral sides divides each of the bases in half. So BM = MC and AN = ND. Further, since a circle can be inscribed in the trapezoids ABMN and NMCD, then
BM + AN = AB + MN,
MC + ND = CD + MN.
It follows that AB = CD, that is, the trapezoid ABCD is isosceles. The desired ratio of areas does not depend on the scale, so we can assume that KN = x, KM = 1. From the right triangles AKN and BKM, we obtain that Rewriting the relation already used above
BM + AN = AB + MN ⇔

We need to calculate the ratio:

Here we have used the fact that the areas of triangles AKD and BKC are related as the squares of the sides KN and KM, i.e. as x2.

Answer:

Task 5. In a convex quadrilateral ABCD, points E, F, H, G are the midpoints of the sides AB, BC, CD, DA, respectively, and O is the intersection point of the segments EH and FG. It is known that EH = a, FG = b, Find the lengths of the diagonals of the quadrilateral.

Solution. It is known that if you connect in series the midpoints of the sides of an arbitrary quadrilateral, you get a parallelogram. In our case EFHG is a parallelogram and O is the intersection point of its diagonals. Then

Apply the cosine theorem to the triangle FOH:

Since FH is the midline of triangle BCD, then

Similarly, applying the cosine theorem to triangle EFO, we get that

Answer:

Task 6. The sides of a trapezoid are 3 and 5. It is known that a circle can be inscribed in a trapezoid. The midline of a trapezoid divides it into two parts, the ratio of the areas of which is equal to Find the bases of the trapezoid.

Solution. Let ABCD be a given trapezoid, AB = 3 and CD = 5 - its sides, points K and M - midpoints of sides AB and CD, respectively. Let, for definiteness, AD > BC, then the area of ​​the trapezoid AKMD will be greater than the area of ​​the trapezoid KBCM. Since KM is the midline of the trapezoid ABCD, the trapezoids AKMD and KBCM have equal heights. Since the area of ​​a trapezoid is equal to the product of half the sum of the bases and the height, then the following equality is true:

Further, since a circle can be inscribed in the trapezoid ABCD, then AD + BC = AB + CD = 8. Then KM = 4 as the midline of the trapezoid ABCD. Let BC = x, then AD = 8 - x. We have:
So BC = 1 and AD = 7.

Answer: 1 and 7.

Task 7. Base AB of trapezoid ABCD is twice as long as base CD and twice as long as lateral side AD. The length of the diagonal AC is a, and the length of the lateral side BC is equal to b. Find the area of ​​the trapezoid.

Solution. Let E be the point of intersection of the extensions of the sides of the trapezoid and CD = x, then AD = x, AB = 2x. Segment CD is parallel to segment AB and twice as short, so CD is the midline of triangle ABE. Therefore, CE = BC = b and DE = AD = x, whence AE = 2x. So triangle ABE is isosceles (AB = AE) and AC is its median. Therefore, AC is also the height of this triangle, and hence

Since triangle DEC is similar to triangle AEB with similarity coefficient, then

Answer:

Task 8. The diagonals of the trapezoid ABCD intersect at point E. Find the area of ​​the triangle BCE if the lengths of the bases of the trapezoid are AB = 30, DC = 24, the side lengths AD = 3 and the angle DAB is 60°.

Solution. Let DH be the height of the trapezoid. From triangle ADH we find that

Since the height of triangle ABC, dropped from vertex C, is equal to the height DH of the trapezoid, we have:

Answer:

Task 9. In a trapezoid, the midline is 4, and the angles at one of the bases are 40° and 50°. Find the bases of the trapezoid if the segment connecting the midpoints of the bases is 1.

Solution. Let ABCD be a given trapezoid, AB and CD its bases (AB< CD), M, N - середины AB и CD соответственно. Пусть также ∠ADC = 50°, ∠BCD = 40°. Средняя линия трапеции равна полусумме оснований, поэтому
AB + CD = 8. Let's extend sides DA and CB to the intersection at point E. Consider triangle ABE, where ∠EAB = 50°. ∠EBA = 40°,
hence ∠AEB = 90°. The median EM of this triangle, drawn from the vertex of the right angle, is equal to half of the hypotenuse: EM = AM. Let EM = x, then AM = x, DN = 4 – x. According to the condition of the problem MN = 1, therefore,
EN = x + 1. From the similarity of triangles AEM and DEN we have:

This means that AB = 3 and CD = 5.

Answer: 3 and 5.

Task 10. A convex quadrilateral ABCD is circumscribed about a circle centered at the point O, while AO = OC = 1, BO = OD = 2. Find the perimeter of quadrilateral ABCD.

Solution. Let K, L, M, N be the tangency points of the circle with the sides AB, BC, CD, DA, respectively, r - the radius of the circle. Since the tangent to the circle is perpendicular to the radius drawn to the point of contact, the triangles AKO, BKO, BLO, CLO, CMO, DMO, DNO, ANO are right-angled. Applying the Pythagorean theorem to these triangles, we get that

Therefore, AB = BC = CD = DA, that is, ABCD is a rhombus. The diagonals of the rhombus are perpendicular to each other, and the point of their intersection is the center of the inscribed circle. From here we easily find that the side of the rhombus is equal and, therefore, the perimeter of the rhombus is equal to

Answer:

Tasks for independent solution

C-1. An isosceles trapezoid ABCD is circumscribed about a circle of radius r. Let E and K be the tangency points of this circle with the sides of the trapezium. The angle between the base AB and the side AD of the trapezoid is 60°. Prove that EK is parallel to AB and find the area of ​​the trapezoid ABEK.
C-2. In a trapezoid, the diagonals are 3 and 5, and the segment connecting the midpoints of the bases is 2. Find the area of ​​the trapezoid.
C-3. Is it possible to circumscribe a circle around quadrilateral ABCD if ∠ADC = 30°, AB = 3, BC = 4, AC = 6?
C-4. In the trapezoid ABCD (AB is the base), the values ​​of the angles DAB, BCD, ADC, ABD and ADB form an arithmetic progression (in the order in which they are written). Find the distance from vertex C to diagonal BD if the height of the trapezoid is h.
C-5. Given an isosceles trapezoid into which a circle is inscribed and around which a circle is circumscribed. The ratio of the height of the trapezoid to the radius of the circumscribed circle is Find the angles of the trapezoid.
C-6. The area of ​​the rectangle ABCD is 48, and the length of the diagonal is 10. On the plane in which the rectangle is located, a point O is chosen so that OB = OD = 13. Find the distance from the point O to the vertex of the rectangle furthest from it.
C-7. The perimeter of parallelogram ABCD is 26. The angle ABC is 120°. The radius of a circle inscribed in triangle BCD is Find the lengths of the sides of the parallelogram if it is known that AD > AB.
C-8. Quadrilateral ABCD is inscribed in a circle centered at point O. Radius OA is perpendicular to radius OB, and radius OC is perpendicular to radius OD. The length of the perpendicular dropped from point C to line AD is 9. The length of segment BC is half the length of segment AD. Find the area of ​​triangle AOB.
C-9. In a convex quadrilateral ABCD, vertices A and C are opposite, and the length of side AB is 3. Angle ABC is angle BCD is Find the length of side AD if you know that the area of ​​the quadrilateral is

C-10. Convex quadrilateral ABCD has diagonals AC and BD. It is known that
AD = 2, ∠ABD = ∠ACD = 90°, and the distance between the point of intersection of the bisectors of triangle ABD and the point of intersection of the bisectors of triangle ACD is Find the length of side BC.
C-11. Let M be the intersection point of the diagonals of a convex quadrilateral ABCD, in which the sides AB, AD and BC are equal. Find the angle CMD if it is known that DM = MC,
and ∠CAB ≠ ∠DBA.
C-12. In quadrilateral ABCD, we know that ∠A = 74°, ∠D = 120°. Find the angle between the bisectors of angles B and C.
C-13. A circle can be inscribed in quadrilateral ABCD. Let K be the point of intersection of its diagonals. It is known that AB > BC > KC, and the perimeter and area of ​​the triangle BKC are 14 and 7, respectively. Find DC.
C-14. In a trapezoid circumscribed about a circle, it is known that BC AD, AB = CD, ∠BAD =
= 45°. Find AB if the area of ​​trapezoid ABCD is 10.
C-15. In the trapezoid ABCD with bases AB and CD, it is known that ∠CAB = 2∠DBA. Find the area of ​​the trapezoid.
C-16. In parallelogram ABCD we know that AC = a, ∠CAB = 60°. Find the area of ​​the parallelogram.
S-17. In quadrilateral ABCD, diagonals AC and BD intersect at point K. Points L and M are respectively the midpoints of sides BC and AD. Segment LM contains point K. Quadrilateral ABCD is such that a circle can be inscribed in it. Find the radius of this circle if AB=3 and LK:KM=1:3.
C-18. Convex quadrilateral ABCD has diagonals AC and BD. In this case, ∠BAC =
= ∠BDC, and the area of ​​the circle circumscribed about the triangle BDC is equal to
a) Find the radius of the circle circumscribed about the triangle ABC.
b) Knowing that BC = 3, AC = 4, ∠BAD = 90°, find the area of ​​quadrilateral ABCD.

Formula for the area of ​​a parallelogram

The area of ​​a parallelogram is equal to the product of its side and the height lowered to this side.

Proof

If the parallelogram is a rectangle, then the equality is satisfied by the rectangle area theorem. Further, we assume that the corners of the parallelogram are not right.

Let $\angle BAD$ be an acute angle in a parallelogram $ABCD$ and $AD > AB$. Otherwise, we will rename the vertices. Then the height $BH$ from the vertex $B$ to the line $AD$ falls on the side $AD$, since the leg $AH$ is shorter than the hypotenuse $AB$, and $AB< AD$. Основание $K$ высоты $CK$ из точки $C$ на прямую $AB$ лежит на продолжении отрезка $AD$ за точку $D$, так как угол $\angle BAD$ острый, а значит $\angle CDA$ тупой. Вследствие параллельности прямых $BA$ и $CD$ $\angle BAH = \angle CDK$. В параллелограмме противоположные стороны равны, следовательно, по стороне и двум углам, треугольники $\triangle ABH = \triangle DCK$ равны.

Let's compare the area of ​​the parallelogram $ABCD$ and the area of ​​the rectangle $HBCK$. The area of ​​the parallelogram is greater by the area $\triangle ABH$, but less by the area $\triangle DCK$. Since these triangles are congruent, their areas are also congruent. This means that the area of ​​a parallelogram is equal to the area of ​​a rectangle with sides long to the side and the height of the parallelogram.

Formula for the area of ​​a parallelogram in terms of sides and sine

The area of ​​a parallelogram is equal to the product of adjacent sides and the sine of the angle between them.

Proof

The height of the parallelogram $ABCD$ lowered to the side $AB$ is equal to the product of the segment $BC$ and the sine of the angle $\angle ABC$. It remains to apply the previous assertion.

Formula for the area of ​​a parallelogram in terms of diagonals

The area of ​​a parallelogram is equal to half the product of the diagonals and the sine of the angle between them.

Proof

Let the diagonals of the parallelogram $ABCD$ intersect at the point $O$ at an angle $\alpha$. Then $AO=OC$ and $BO=OD$ by the parallelogram property. The sines of the angles that add up to $180^\circ$ are $\angle AOB = \angle COD = 180^\circ - \angle BOC = 180^\circ - \angle AOD$. Hence, the sines of the angles at the intersection of the diagonals are equal to $\sin \alpha$.

$S_(ABCD)=S_(\triangle AOB) + S_(\triangle BOC) + S_(\triangle COD) + S_(\triangle AOD)$

according to the axiom of area measurement. Apply the triangle area formula $S_(ABC) = \dfrac(1)(2) \cdot AB \cdot BC \sin \angle ABC$ for these triangles and angles when the diagonals intersect. The sides of each are equal to half the diagonals, the sines are also equal. Therefore, the areas of all four triangles are $S = \dfrac(1)(2) \cdot \dfrac(AC)(2) \cdot \dfrac(BD)(2) \cdot \sin \alpha = \dfrac(AC \ cdot BD)(8) \sin \alpha$. Summing up all the above, we get

$S_(ABCD) = 4S = 4 \cdot \dfrac(AC \cdot BD)(8) \sin \alpha = \dfrac(AC \cdot BD \cdot \sin \alpha)(2)$