Quest Source: Decision 5346.-13. OGE 2016 Mathematics, I.V. Yashchenko. 36 options.
Task 11. In trapezoid ABCD, we know that AB = CD, angle BDA = 54°, and angle BDC = 33°. Find angle ABD. Give your answer in degrees.
Decision.
Given an isosceles trapezoid with sides AB=CD. Since the angles at the bases of such a trapezoid are equal, we have that and . Let's find the value of angles A and D. It can be seen from the figure that angle D (and hence angle A) is equal to:
Now consider triangle ABD, in which angles A and BDA are known, and since the sum of all angles in the triangle is 180 degrees, we find the third angle ABD:
Answer: 39.
Task 12. Three points are marked on 1x1 checkered paper: A, B and C. Find the distance from point A to line BC.
Decision.
The distance from point A to line BC is the normal dropped from point A to side BC (red line in the figure). The length of this normal is 3 cells, that is, 3 units.
Answer: 3.
Task 13. Which of the following statements are correct?
1) The area of a triangle is less than the product of its two sides.
2) The angle inscribed in a circle is equal to the corresponding central angle based on the same arc.
3) Through a point not lying on a given line, one can draw a line perpendicular to this line.
Decision.
1) True. The area of a triangle is equal to the product of the height and half the base of the triangle, and all these quantities are less than the lengths of any two of its sides.
Theorem 1 (Thales theorem). Parallel lines cut proportional segments on the lines intersecting them (Fig. 1).
Definition 1 . Two triangles (Fig. 2) are called similar if their corresponding sides are proportional.
Theorem 2 (first sign of similarity). If the angle of the first triangle is equal to the angle of the second triangle, and the sides of the triangles adjacent to these angles are proportional, then such triangles are similar (see Fig. 2).
Theorem 3 (second sign of similarity). If two angles of one triangle are equal, respectively, to two angles of another triangle, then such triangles are similar (Fig. 3).
Theorem 4 (Menelaus' theorem). If some line intersects sides AB and BC of triangle ABC at points X and Y, respectively, and the continuation of side AC is at point Z (Fig. 4), then
Theorem 5. Let heights AA1 and CC1 be drawn in an acute-angled triangle ABC (Fig. 5). Then the triangles A1 BC1 and ABC are similar, and the similarity coefficient is equal to cos ∠B.
Lemma 1. If sides AC and DF of triangles ABC and DEF lie on the same line or on parallel lines (Fig. 6), then
Lemma 2. If two triangles have a common side AC (Fig. 7), then
Lemma 3. If triangles ABC and AB1 C1 have a common angle A, then
Lemma 4. The areas of similar triangles are related as the square of the similarity coefficient.
Proofs of some theorems
Proof of Theorem 4
. Draw a line through point C parallel to line AB until it intersects line XZ at point K (Fig. 9). We have to prove that
Consider two pairs of similar triangles:
Multiplying these equalities term by term, we get: 
Q.E.D.
Proof of Theorem 5. Let us prove the similarity of triangles A1 BC1 and ABC using the first similarity test. Since these two triangles have a common angle B, it suffices to prove that
But this follows from the fact that from the right triangle ABA1, but from the right triangle CBC1. Along the way, the second part of the theorem is also proved.
Problem solving
Task 1. Given a trapezoid ABCD, and it is known that BC = a and AD = b. Parallel to its bases BC and AD, a straight line is drawn that intersects side AB at point P, diagonal AC at point L, diagonal BD at point R, and side CD at point Q (Fig. 10). It is known that PL = LR. Find P.Q.
Decision. Let us first prove that PL = RQ. Consider two pairs of similar triangles: 
According to the Thales theorem, we have:
Let us now denote PL = LR = RQ = x and consider again two pairs of similar triangles: 
We have next:
Means,
Answer:
Task 2. In triangle ABC, angle A is 45° and angle C is acute. From the midpoint N of side BC the perpendicular NM is dropped to side AC (Fig. 11). The areas of triangles NMC and ABC are related respectively as 1: 8. Find the angles of triangle ABC.
Decision. Let BH be the height dropped from vertex B to side AC.
Since NM is the midline of triangle BHC, then S∆BHC = 4S∆NMC .
But, according to the condition of the problem, S∆ABC = 8S∆NMC .
Therefore, S∆ABC = 2S∆BHC , so S∆ABH = S∆BHC . So AH = HC,
whence ∠CAB = ∠ACB = 45°, ∠ABC = 90°.
Answer: ∠CAB = ∠ACB = 45°, ∠ABC = 90°.
Task 3. Given a triangle ABC in which angle B is equal to 30°, AB = 4 and BC = 6. The bisector of angle B intersects side AC at point D (Fig. 12). Find the area of triangle ABD.
Decision. Let's apply the interior angle bisector theorem to triangle ABC:
Means,
Answer:
The article was published with the support of the World of Flowers company. Wholesale and retail warehouse of wedding and ritual goods, artificial flowers in Krasnodar. Wedding accessories - candles, posters, glasses, ribbons, invitations and more. Ritual goods - fabrics, clothes, accessories. You can learn more about the company, view the product catalog, prices and contacts on the website, which is located at: flowersworld.su.
Task 4. Through the midpoint M of the side BC of the parallelogram ABCD, whose area is 1, and the vertex A, a line is drawn that intersects the diagonal BD at the point O (Fig. 13). Find the area of quadrilateral OMCD.
Decision. We will search for the area of the quadrilateral OMCD as the difference between the areas of triangles BCD and BOM. The area of triangle BCD is equal to half the area of parallelogram ABCD and is equal to Find the area of triangle BOM. We have:
∆ BOM ∼ ∆ AOD ⇒ 
Further:
Means, 
Answer:
Task 5. A right-angled triangle MNC is inscribed in a right-angled isosceles triangle ABC with a right angle at vertex B so that angle MNC is right, point N lies on AC, and point M lies on side AB (Fig. 14). In what ratio must point N divide the hypotenuse AC so that the area of triangle MNC is equal to the area of triangle ABC?
Decision. We can assume that AB = 1. Denote AM = x, 0< x < 1, тогда BM = 1 – x,
We have: 
Answer:
Task 6. In trapezoid ABCD, diagonal AC is perpendicular to side CD, and diagonal DB is perpendicular to side AB. The extensions of the sides AB and DC intersect at the point K, forming a triangle AKD with an angle of 45° at the apex K (Fig. 15). The area of trapezoid ABCD is equal to S. Find the area of triangle AKD.
Decision. According to Theorem 5, triangle BKC is similar to triangle AKD with similarity coefficient 
Therefore, the areas of these triangles are in ratio as 1:2, which means that the area of trapezoid ABCD is equal to the area of triangle BKC. Therefore, the area of triangle AKD is 2S.
Answer: 2S.
Task 7. In triangle ABC, point K is taken on side AB so that AK: KB = 1: 2, and point L is taken on side BC so that CL: LB = 2: 1. Let Q be the intersection point of lines AL and CK (Fig. sixteen). Find the area of triangle ABC knowing that the area of triangle BQC is 1.
Decision. Let AK = x, BL = y. Then KB = 2x,
LC = 2y, so AB = 3x and BC = 3y. Let us apply the Menelaus theorem to the triangle ABL and the secant KQ:
Task 8. From the point M, which is located inside the acute-angled triangle ABC, perpendiculars are dropped to the sides (Fig. 17). The lengths of the sides and the perpendiculars dropped on them, respectively, are equal a and k, b and m, c and n. Calculate the ratio of the area of triangle ABC to the area of a triangle whose vertices are the bases of the perpendiculars.
Decision. We introduce standard notation, that is, we denote the lengths of the sides of the triangle ABC: BC = a, CA = b, AB = c; angles: ∠BAC = α,
∠ABC = β, ∠ACB = γ. The bases of the perpendiculars dropped from the point M to the sides BC, CA, and AB will be denoted by D, E, and F, respectively. Then, according to the condition of the problem, MD = k, ME = m, MF = n. It is obvious that the angle EMF is equal to π - α, the angle DMF is equal to π - β, the angle DME is equal to π - γ and the point M is located inside the triangle DEF. The area of triangle DEF is:
The area of triangle ABC is:
Find the ratio of the areas of triangles DEF and ABC:
Hence, 
Answer: 
Task 9. Points P and Q are located on side BC of triangle ABC so that BP: PQ: QC = 1:2:3.
Point R divides side AC of this triangle in such a way that AR: RC = 1: 2 (Fig. 18). What is the ratio of the area of quadrilateral PQST to the area of triangle ABC, where S and T are the points of intersection of line BR with lines AQ and AP, respectively?
Decision. Denote BP = x, AR = y; then
PQ=2x, QC=3x, RC=2y. Let us calculate what part of the area of the quadrilateral PQST is the area of the triangle APQ, and hence the area of the triangle ABC. To do this, we need relations in which the points S and T divide the lines AQ and AP, respectively. Let us apply the Menelaus theorem to the triangle ACQ and the secant SR:
Similarly, applying the Menelaus theorem to the triangle ACP and the secant TR, we get:
Further: 
On the other hand, applying the area lemma to triangles APQ and ABC, we get
Answer:
Task 10. In triangle ABC, the length of height BD is equal to 6, the length of median CE is equal to 5, the distance from the point of intersection of BD with CE to side AC is equal to 1 (Fig. 19). Find the length of side AB.
Decision. Let point O be the point of intersection of lines BD and CE. The distance from the point O to the side AC (which is equal to one) is the length of the segment OD. So, OD = 1 and OB = 5. Apply Menelaus' theorem to the triangle ABD and the secant OE:
Applying now the theorem of Menelaus to the triangle ACE and the secant OD, we get that
whence OE = 2CO, and taking into account OE + CO = CE = 5
we get that We apply the Pythagorean theorem to the right triangle CDO:
Means, 
Finally, consider a right triangle ABD, in which we also use the Pythagorean theorem:
Answer:
Task 11. Points C and D lie on segment AB, and point C is between points A and D. Point M is taken so that lines AM and MD are perpendicular, and lines CM and MB are also perpendicular (Fig. 20). Find the area of triangle AMB if the angle CMD is known to be α and the areas of triangles AMD and CMB are S1 and S2, respectively.
Decision. Denote the areas of triangles AMB and CMD, respectively, by
x and y (x > y). Note that x + y = S1 + S2 . Let us now show that xy = S 1 S 2 sin 2 α. Really, 
Likewise, 
Since ∠AMB = ∠AMC + ∠CMD + ∠DMB =
= 90° – α + α + 90° – α = 180° – α, and sin ∠AMB =
= sinα. Means:
Thus the numbers x and y are the roots of the quadratic equation
t2 – (S1 + S2 )t + S1 S2 sin2 α = 0.
The larger root of this equation is:
Answer: 
Tasks for independent solution
C-1. In a triangle ABC whose area is S, the bisector CE and the median BD are drawn, intersecting at point O. Find the area of quadrilateral ADOE, knowing that BC = a, AC = b.
C-2. A square is inscribed in an isosceles triangle ABC so that two of its vertices lie on the base of BC, and the other two lie on the sides of the triangle. The side of a square is related to the radius of a circle inscribed in a triangle, as
8:5. Find the corners of the triangle.
C-3. In parallelogram ABCD with sides AD = 5 and AB = 4, a line segment EF is drawn connecting point E of side BC with point F of side CD. Points E and F are chosen so that
BE: EC = 1: 2, CF: FE = 1: 5. It is known that the intersection point M of the diagonal AC with the segment FE satisfies the condition MF: ME = 1: 4. Find the diagonals of the parallelogram.
C-4. The area of trapezoid ABCD is equal to 6. Let E be the point of intersection of the extensions of the sides of this trapezoid. Through the point E and the point of intersection of the diagonals of the trapezoid, a straight line is drawn that intersects the smaller base BC at the point P, the larger base AD - at the point Q. The point F lies on the segment EC, and EF: FC = EP: EQ = 1: 3.
Find the area of triangle EPF.
C-5. In an acute-angled triangle ABC (where AB > BC) heights AM and CN are drawn, point O is the center of the circle circumscribed about triangle ABC. It is known that the magnitude of angle ABC is β, and the area of quadrilateral NOMB is S. Find the length of side AC.
C-6. In triangle ABC, point K on side AB and point M on side AC are located in such a way that the relations AK: KB = 3: 2 and AM: MC = 4: 5 hold. In what ratio does the intersection point of lines KC and BM divide segment BM?
C-7. Point D is taken inside a right-angled triangle ABC (angle B is right) so that the areas of triangles ABD and BDC are respectively three and four times less than the area of triangle ABC. The lengths of segments AD and DC are equal to a and c, respectively. Find the length of segment BD.
S-8. In a convex quadrilateral ABCD on side CD, a point E is taken so that the segment AE divides quadrilateral ABCD into a rhombus and an isosceles triangle, the ratio of the areas of which is equal to Find the value of the angle BAD.
C-9. The height of the trapezoid ABCD is 7, and the lengths of the bases AD and BC are 8 and 6, respectively. Through the point E, lying on the side CD, a line BE is drawn, which divides the diagonal AC at the point O in relation to AO: OC = 3: 2. Find the area triangle OEC.
S-10. Points K, L, M divide the sides of the convex quadrilateral ABCD with respect to AK: BK = CL: BL = CM: DM = 1: 2. It is known that the radius of the circle circumscribed about the triangle KLM is KL = 4, LM = 3 and KM< KL. Найдите площадь четырехугольника ABCD.
S-11. The extensions of sides AD and BC of a convex quadrilateral ABCD intersect at point M, and the extensions of sides AB and CD intersect at point O. Segment MO is perpendicular to the bisector of angle AOD. Find the area ratio of triangles AOD and BOC if OA = 6, OD = 4, CD = 1.
S-12. In triangle ABC, the angle at vertex A is 30°, and heights BD and CE intersect at point O. Find the ratio of the radii of the circumcircles of triangles DEO and ABC.
S-13. The segments connecting the bases of the altitudes of an acute-angled triangle are 5, 12 and 13. Find the radius of the circle circumscribed about the triangle.
S-14. In an acute-angled triangle ABC, point M is taken at height AD, and point N is taken at height BP, so that angles BMC and ANC are right. The distance between points M and N is a ∠MCN = 30°.
Find the bisector CL of triangle CMN.
S-15. Points D, E and F are taken on sides AB, BC and AC of triangle ABC, respectively. Segments AE and DF pass through the center of a circle inscribed in triangle ABC, and lines DF and BC are parallel. Find the length of segment BE and the perimeter of triangle ABC if BC = 15, BD = 6, CF = 4.
S-16. In triangle ABC, the bisector BB" intersects the median AA" at point O.
Find the ratio of the area of triangle BOA" to the area of triangle AOB" if AB:AC = 1:4.
S-17. In triangle ABC, point D lies on AC, and AD = 2DC. Point E lies on BC. The area of triangle ABD is 3, the area of triangle AED is 1. Segments AE and BD intersect at point O. Find the ratio of the areas of triangles ABO and OED.
S-18. In the parallelogram ABCD, the points E and F lie respectively on the sides AB and BC, M is the point of intersection of the lines AF and DE, with AE = 2BE and BF = 3CF. Find the ratio AM:MF.
S-19. In rectangle ABCD on sides
AB and AD, points E and F are chosen, respectively, so that AE: EB = 3: 1, AF: FD = 1: 2. Find EO: OD, where O is the intersection point of the segments DE and CF.
S-20. Point N is taken on side PQ of triangle PQR, and point L is taken on side PR, and
NQ=LR. The intersection point of the segments QL and NR divides the segment QL in the ratio m: n, counting from the point Q. Find the ratio PN: PR.
S-21. Points A and B are taken on the sides of an acute angle with vertex O. On ray OB, point M is taken at a distance of 3OA from line OA, and on ray OA, point N is taken at a distance of 3OB from line OB. The radius of the circumcircle of triangle AOB is 3. Find MN.
S-22. In a convex pentagon ABCDE, the diagonals BE and CE are the bisectors of the vertex angles B and C, respectively, ∠A = 35°, ∠D = 145°, S∆BCE = 11. Find the area of the pentagon ABCDE.
S-23. On the bases AD and BC of the trapezoid ABCD, squares ADEF and BCGH are constructed, located outside the trapezoid. The diagonals of the trapezoid intersect at point O. Find the length of segment AD if BC = 2, GO = 7, and GF = 18.
S-24. In triangle ABC we know that AB = BC and angle BAC is 45°. Line MN intersects side AC at point M and side BC at point N, with AM = 2MC and ∠NMC = 60°. Find the ratio of the area of triangle MNC to the area of quadrilateral ABNM.
S-25. In triangle ABC, point N is taken on side AB, and point M is taken on side AC. Segments CN and BM intersect at point O, AN: NB = 2: 3,
BO: OM = 5: 2. Find CO: ON.
Trapeze on the exam. A basic level of.
Tasks from the open bank of FIPI tasks.
Task 1.In the trapezoid ABCD, we know that AB=CD,∠ BDA=54° and ∠ BDC=23°. Find angle ABD. Give your answer in degrees.
Decision.In this trapezoid, angle A DC at the lower base is equal to the sum of the angles A D V and V DC , is equal to 54 + 23 = 77 degrees. Since the trapezoid is isosceles, the angles at the lower base are equal and the angle BA D is also 77 degrees. Sum of angles VA D and AB D equal to 180 degrees (one-sided with parallel lines A D and BC and secant AB). So the angle ABC is equal to 180 - 77 \u003d 103 degrees.
Next, we use the equality of angles A D B and D BC (cross-lying with parallel lines A D and BC and secant B D). So the angle AB D equal to 103 - 54 \u003d 49 degrees.
Answer 49.
Task 2.The bases of an isosceles trapezoid are 10 and 24, the side is 25. Find the height of the trapezoid.
Decision.In this trapezoid, the upper base BC is 10, the lower A D =24. From vertices B and C we lower the heights to the lower base. In the resulting rectangle NVSK NK=BC=10. Triangles ABH and K DC DC ), so AH \u003d K D =(24-10):2=7. According to the Pythagorean theorem, in a triangle ABN, the square of the leg BH is equal to the difference between the square of the hypotenuse AB and the square of the leg AN. That is, VN 2 \u003d 625 - 49 \u003d 576. VN \u003d 24.
Answer 24.
Task 3.In an isosceles trapezoid, one of the bases
is 3 and the other is 7. The height of the trapezoid is 4. Find the tangent of the acute angle of the trapezoid.
Decision.In this trapezoid, the upper base BC is 3, the lower A D =7. From vertices B and C we lower the heights to the lower base. In the resulting rectangle NVSK NK=BC=3. Triangles ABH and K DC are equal (they are rectangular, BH = SK, AB = DC ), so AH \u003d K D =(7-3):2=2. The tangent of an acute angle BAN in a right triangle ABN is equal to the ratio of the opposite leg BH to the adjacent leg AH, that is, 4:2=2.
Answer 2.
Task 4.The bases of the trapezoid are 8 and 16, the lateral side, equal to 6, forms an angle of 150 ° with one of the bases of the trapezoid. Find the area of the trapezoid.
Decision.Let in the trapezoid in the figure of the base BC \u003d 8, AD =16, side AB=6, and angle ABC is 150 degrees. We know that the area of a trapezoid is equal to the product of half the sum of the bases and the height. The bases are known. Let's find the height of BH. In a right triangle ABH, the angle ABH is 150 - 90 = 60 degrees. So the angle VAN is equal to 90 - 60 \u003d 30 degrees. And in a right triangle, the leg opposite the angle of 30 degrees is equal to half the hypotenuse. So VN=3.
It remains to calculate the area of the trapezoid. The half sum of the bases is equal to (8+16):2=12. The area is 12*3=36.
Answer 36.
Task 5.In a rectangular trapezoidABCD with grounds Sun and BUTD injection ATAD straight, AB=3, Sun=CD=5. Find the midline of the trapezoid.
Decision.The median line of the trapezoid is half the sum of the bases. In this trapezoid, the upper base BC is 5, the lower A D unknown. From vertex C we lower the height to the lower base. In the resulting rectangle NVSK AH=BC=5, CH=AB=3. Triangle H DC rectangular. By the Pythagorean theorem, the square of the leg H D equal to the difference of the square of the hypotenuse DC and the square of the leg CH. That is, N D 2 \u003d 65 -9 \u003d 16. H D \u003d 4. So the lower base A D =AH+H D =5+4=9. The median line of the trapezoid is (5+9):2=7.
Answer 7.
Task 6.In a rectangular trapezoid, the bases are 4 and 7, and one of the angles is 135°. Find the smaller side.
Decision.Let's use the drawing for the previous problem. In this trapezoid, the upper base BC is 4, the lower A D=7. Angle BC D is equal to 135 degrees. From vertex C we lower the height to the lower base. Then H D =7-4=3. In the resulting right triangle H DC angle HC D equals 135-90=45 degrees. So the angle H DC also 45 degrees. Legs CH= H D=3.
Answer 3.
Tasks for independent solution.
- ∠ BDA=40° and ∠ BDC=30°. Find angle ABD. Give your answer in degrees.
- in a trapeze ABCD it is known that AB=CD, ∠ BDA=45° and ∠ bdc=23°. Find an angle ABD. Give your answer in degrees.
- In the trapezoid ABCD, we know that AB=CD,∠ BDA=49° and ∠ BDC=31°. Find angle ABD. Give your answer in degrees.
- The bases of an isosceles trapezoid are 7 and 13, the side is 5. Find the height of the trapezoid.
- The bases of an isosceles trapezoid are 11 and 21, the side is 13. Find the height of the trapezoid.
- The bases of the trapezoid are 10 and 20, the lateral side, equal to 8, forms an angle of 150 ° with one of the bases of the trapezoid. Find the area of the trapezoid.
- In an isosceles trapezoid, one of the bases is 5 and the other is 9. The height of the trapezoid is 6. Find the tangent of the acute angle of the trapezoid.
- In a rectangular trapezoidABCD with grounds Sun and BUTD injection ATAD straight, AB=8, Sun=CD=10. Find the midline of the trapezoid.
- In a rectangular trapezoidABC D with grounds Sun and BUT D injection AT AD straight, AB = 15 , Sun = CD = 17 . Find the midline of the trapezoid.
- In a rectangular trapezoid, the bases are 3 and 5, and one of the angles is 135°. Find the smaller side.
