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In △ ABC, D and E are on the lines AB and AC respectively, ad = 2, BD = 5, EA = 4, EC = 10, Ed: BC=
AD/AB=2/(2+5)=2/7
AE/AC=4/(4+10)=4/14=2/7
Ad / AB = AE / AC, so de ‖ BC
AD /AB=DE/BC=2/7
Known: as shown in the figure, in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is ()
A. 30°B. 36°C. 45°D. 50°
Let ∠ EBD = x °, be = De, EDB = ∠ EBD = x °, AED = ∠ EBD + ∠ EDB = 2x °, ad = De, a = ∠ AED = 2x °, BDC = ∠ a + ∠ abd = 3x °, BD = BC, C = ∠ BDC = 3x °, AB = AC, ABC = ∠ C = 3x °, a + ∠ ABC + ∠ C = 180 °, 2x +
RT triangle ABC, ∠ ACB = 90 °, ab = 5cm, AC = 4cm, CD ⊥ AB in D?
In △ ACB and △ CDB, ∠ ABC = ∠ CBD. ∠ CDB = ∠ ACB = 90 ° because the sum of internal angles of triangles is equal, ∠ a = ∠ BCD. So △ ACB is similar to △ CDB. So CB: ab = CD: AC, CB ^ 2 = AB ^ 2-ac ^ 2, that is CB = 3. So 3:5 = CD: 4, the solution is: CD = 12 / 5 (12 / 5)
In RT triangle ABC, the angle c = 90 ° if a: B = 3:4, C = 10, then a =
A=6 B=8
It is known that, as shown in the figure, △ ABC, point E is on the central line BD, ∠ DAE = ∠ abd
It is proved that: (1) ∵ - DAE = - abd, ∵ - ADE = - BDA, ∵ △ ade ∵ BDA. (2 points) ∵ - addd = dead, (2 points) & nbsp; that is, ad2 = de · dB. (1 point) (2) ∵ D is the midpoint on the AC side, ∵ ad = DC. ∵ addd = dead, ∵ dcbd = dedc, (2 points) ∵ - CDE = - BDC. (1 point)
RELATED INFORMATIONS
- 1. It is known that CD is the height of AB side in △ ABC, AC + BC = 8, CD = 2, AE is the diameter of circumscribed circle of △ ABC, (1) try to explain △ CBD ∽ AEC, (2) let AC = x, AE = y, find the functional relation of y with respect to x, and (3) find the maximum value of Y
- 2. As shown in the figure, in equilateral △ ABC, CD intersects be at point F, ∠ BFC = 120?: ad = CE
- 3. In 3-angle ABC, the angle BAC is 90 degrees, ad is perpendicular to BC, EF is perpendicular to BC, FM is perpendicular to AC, the perpendicular foot is D, F, m respectively, angle 1 = angle 2, verification: FM = FD
- 4. As shown in the figure, it is known that CD is perpendicular to ab at point D, be is perpendicular to AC at point E, and CD intersects be at point o (1) If OC = ob, prove that point O is on the bisector of angular BAC (2) If point O is on the bisector of angle BAC, OC = ob is proved
- 5. In the triangle ABC, ab = AC, D is a point on BC, and
- 6. As shown in the figure, in RT △ ABC, ab = AC, ∠ BAC = 90 °, ∠ 1 = ∠ 2, CE ⊥ BD is extended to E. verification: BD = 2ce
- 7. As shown in the figure, in the triangle ABC, the angle ACB = 90 °, CD is perpendicular to AB, the perpendicular foot is D, and the angle a = 60 ° to prove BD = 3aD
- 8. As shown in the figure, in the triangle ABC, the angle ACB is 90 degrees, the angle CD is high, and the angle a is 30 degrees
- 9. In the triangle ABC, we know that the opposite sides of angles a, B and C are a, B and C respectively, and a * a + b * B-C * C = / 3A * B. find the size of angle A
- 10. AB = AC, < BAC = < DAE, ad = AE, BD = CE
- 11. It is known that: as shown in the figure, the circle O with the diameter of AB on one side of the equilateral triangle ABC intersects with the sides AC and BC at points D and e respectively, DF ⊥ BC is made through point D, and F is the perpendicular foot (1) Proof: DF is tangent of circle O (2) If the side length of equilateral triangle ABC is 4, find the length of DF (3) Finding the area of the shadow in a graph The first question is not, the shadow area is DFE
- 12. If the square of a plus the square of B equals twice the square of C, then the minimum value of COSC is
- 13. In the triangle ABC, a = 120 °, ab = 5, BC = 7, then the value of sina / SINB
- 14. In the right triangle ABC, C is equal to 90 degrees, AC is equal to twice BC, then the value of sina is?
- 15. In the right triangle ABC, angle c is right angle, angle B ∶ angle c = 1 ∶ 4, find the degree of angle A The middle equation is still missing
- 16. If the angle c = 90 degrees, (a + b) * (a + b) = 12 and C = 2 in the right triangle ABC, then the s triangle ABC =?
- 17. The three sides of triangle are equal difference series, the circumference is 36, and the circumference of inscribed circle is 6 π Some people do that ∵ the three sides grow into an arithmetic sequence with a circumference of 36 Let the three sides be 12-n, 12, 12 + n respectively The circumference of the inscribed circle is 6 π --- radius of the inscribed circle r = 3 --- triangle area = (36 / 2) * 3 = 54 From Helen's formula: 54 & sup2; = 18 * 6 * (6 + n) * (6-n) --->27=36-n²--->n²=9--->n=3 --->The lengths of the three sides are 9, 12 and 15 respectively ∵ 9 & sup2; + 12 & sup2; = 15 & sup2;, ∵ the triangle is a right triangle Then why must one side be 12?
- 18. Three straight lines PA, Pb, PC, ∠ APC = ∠ APB = 60 ° and ∠ BPC = 90 ° in space are used to calculate the dihedral angle b-pa-c
- 19. As shown in the figure, in △ ABC, ab = AC, e is on AC, and the extension line of ad = AE, de intersects BC at point F
- 20. If alpha is equal to - 3, then which quadrant is the end of alpha?