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As shown in the figure, △ ABC, ab = AC, point E is on the extension line of Ca, and ∠ AEF = ∠ AFE. What is the position relationship between the straight line EF and BC? Why?
EF⊥BC．
The extension EF intersects BC at point D, let ∠ AEF = ∠ AFE = ∠ BFD = X,
∵AB=AC，
∴∠B=∠C，
∵∠B+∠C=∠BAE=180°-2x，
∴∠B=∠C=90°-x，
∴∠BDE=180°-∠B-∠BFD=180°-（90°-x）-x=90°，
∴EF⊥BC．
As shown in the figure, △ ABC, ab = AC, point E is on the extension line of Ca, and ∠ AEF = ∠ AFE. What is the position relationship between the straight line EF and BC? Why?
EF⊥BC．
The extension EF intersects BC at point D, let ∠ AEF = ∠ AFE = ∠ BFD = X,
∵AB=AC，
∴∠B=∠C，
∵∠B+∠C=∠BAE=180°-2x，
∴∠B=∠C=90°-x，
∴∠BDE=180°-∠B-∠BFD=180°-（90°-x）-x=90°，
∴EF⊥BC．
As shown in the figure, △ ABC, ab = AC, point E is on the extension line of Ca, and ∠ AEF = ∠ AFE. What is the position relationship between the straight line EF and BC? Why?
EF⊥BC．
The extension EF intersects BC at point D, let ∠ AEF = ∠ AFE = ∠ BFD = X,
∵AB=AC，
∴∠B=∠C，
∵∠B+∠C=∠BAE=180°-2x，
∴∠B=∠C=90°-x，
∴∠BDE=180°-∠B-∠BFD=180°-（90°-x）-x=90°，
∴EF⊥BC．
As shown in the figure, in triangle ABC, CE bisects angle ACB, CF bisection angle ACD, and ef ∥ BC intersects AC with M. if cm = 5, calculate the square of CM + the square of CF
There is something wrong with the problem, it should be to find the value of CE? + CF? Oh, CE bisection angle ACB, CF bisection angle ACD, BCE = ∠ ECM, ? MCF = ∠ FCD, ∠ ECF = 90 ° and ? EF ∵ BC, ? BCE = ∠ ECM = ∠ mec, ? MCF = ∠ FCD = ∠ MFC, EM = MC = MC = MF = 5, CE ∧ CF ? EF ∫ BC
As shown in the figure, in triangle ABC, D is on the extension line of BC, and AC is equal to CD, CE is the center line of triangle ACD, CF bisector angle ACB, AB was crossed to F, and CE vertical CF and CF parallel AD were verified
∵ CE is the center line of the triangle ACD
∴AE=ED
∵AC=CD CE=CE
All △ AEC is equal to △ Dec
∴∠ACE=∠DCE=∠ACD/2 ∠AEC=∠DEC
∵ CF bisection angle ACB
∴∠ACF=∠ACB/2
∴∠FCE=∠ACE+∠ACF=∠ACD/2+∠ACB/2=∠ACB/2=180/2=90
∴CE⊥CF
∵∠AEC+∠DEC=180
∴∠AEC=∠DEC=90
∴CE⊥AD
∵CE⊥CF
∴AD||CF
As shown in the figure, in △ ABC, CE bisects ∠ ACB to intersect AB to e, and e to be EF ‖ BC to intersect ∠ ACD bisector to f and EF to AC to M. if cm = 5, then CE2 + CF2=______ .
In this paper, we discuss the difference between the ACB and the AB in the E, CF, and ACD, and the CF is ∠ ACD, and ∠ 1 = 1 = 2 = 12 ∠ ACB, and ∠ 3 = 3 = 4 = 4 = 12 ∠ ACD, and ＜ 2 + 2 ∠ 3 = 12 (∵ ACB + ACD) = 90 °, and \△ CEF is a right triangle, ∵ EF 8757; EF ≓ BC, ∵ EF ≓ BC, ∵ 1 ? 1 ? 5, 87878757; 2 ∵ 2 9 5, ᙽ 3 M 3= 5
As shown in the figure, the triangle ABC is an equilateral triangle, D is a point on the extension line BC, CE bisects the angle ACD, CE equals BD Prove that the triangle ade is an equilateral triangle
Because CE bisects the angle ACD, AB is parallel to CE, and the triangle ACF is equilateral triangle, AF = AC, angle AFE = 120 = angle ACD, and because CE = BD, CF = BC, so Fe = CD, so triangle ACD is equal to triangle AFE, so ad = AE, angle CAD = angle FAE, so angle CAF = angle DAE = 60, so
It is known that △ ABC is an isosceles right triangle with a right angle side length of 1. Take the hypotenuse AC of RT △ ABC as the right angle side, draw the second isosceles RT △ ACD, and then draw the third isosceles RT △ ade The length of the hypotenuse of the nth isosceles right triangle is______ .
According to the Pythagorean theorem, the length of the hypotenuse of the first isosceles right triangle is
The length of the hypotenuse of the second isosceles right triangle is 2=（
2) The length of the hypotenuse of the third isosceles right triangle is 2
2=（
2) The length of the hypotenuse of the nth isosceles right triangle is 3（
2）n．
As shown in the figure, it is known that RT △ ABC is an isosceles right triangle with a right angle side length of 1 As shown in the figure, It is known that RT △ ABC is an isosceles right triangle with a right angle side length of 1, Let the hypotenuse AC of RT △ ABC be the right angle side, Draw the second isosceles RT △ ACD, Then take the oblique side ad of RT △ ACD as the right angle side, Draw the third isosceles RT △ ade and so on The length of the hypotenuse of the 2n isosceles right triangle is____________ If the picture, you can draw it on the paper, or I will draw a picture immediately.
The 2N-1 power of radical 2
As shown in the figure, given that the right angle side length of isosceles RT △ ABC is 1, draw the second isosceles RT Δ ACD with the oblique side AC of RT △ ABC as the right angle side, and then draw the third isosceles RT Δ ade By analogy, until the fifth isosceles RT △ AFG, the image area formed by these five isosceles right triangles is—— I need it before 13:45 this afternoon. Come on, if it's right, it will pay off
1 / 2 + 1 + 2 + 4 + 8 = 15 and 1 / 2
RELATED INFORMATIONS
- 1. In the isosceles right triangle ABC, the angle ABC = 90 degrees D is the midpoint on the edge of AC, where de is perpendicular to DF and intersects AB at If AE = 4, FD = 3, find the length of EF
- 2. It is known that in isosceles RT △ ABC, ∠ a = 90 ° D is the midpoint of BC, e and F are points on AB and AC respectively, and EA = CF is satisfied. Verification: de = DF
- 3. It is known that: as shown in the figure, ∠ ACB = 90 ° in △ ABC, CD is high, CE bisects ∠ BCD, and ∠ ACD: ∠ BCD = 1:2, then CE is the center line on AB side, right? Give reasons
- 4. It is known that in △ ABC, ab = AC, CD ⊥ ad is in D, CD = 1 2BC, D is outside △ ABC, and it is proved that ∠ ACD = ∠ B
- 5. Both the triangle ABC and the triangle ECD are isosceles right triangles. The angle ACB = angle DCE = 90 ° D is a point on the edge of AB and connected with Ae. When de = 17 and AE = 15, the triangle ABC and ECD are isosceles right triangles, Find the length of ad
- 6. If there is a point D in the triangle ABC that connects Da, DB and DC, there are () triangles that do not overlap each other
- 7. As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, AC = BC, ∠ CAD = ∠ bad, AB = AC + CD
- 8. In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and Tanc = 3 7． (I) find the value of COSC; (II) if CB• CA=5 2, and a + B = 9, find the length of C
- 9. In the RT triangle ABC, ∠ C = 90 ° a, B, C denote the opposite sides of ∠ a, ∠ B and ∠ C respectively. Given that a = 6 under the radical sign, ∠ a = 60 ° find B, C
- 10. It is known that in RT △ ABC ∠ ACB = 90 ° CD ⊥ AB, the perpendicular foot is d BC = 2, BD = radical 3, and the acute angles in △ ABC △ ACDC △ BCD are calculated respectively
- 11. As shown in the figure, △ ABC, D is on the extension line of BC, and AC = CD, CE is the middle line of △ ACD, and CF bisects ∠ ACB As shown in the figure, △ ABC, D is on the extension line of BC, and AC = CD, CE is the center line of △ ACD, CF bisects ∠ ACB, intersects AB and F, and proves (1) ce ⊥ CF: (2) CF is parallel to AD
- 12. In the right triangle ABC, the angle c = 90 degrees, AC = 4, BC = 3, ab = 5, P is the intersection point of the bisector of the inner angle of the triangle ABC. Find the distance between P and each side
- 13. In the triangle ABC, angle B = 90 °, ab = 6, BC = 3, point P moves from point a to side AB at a speed of 1cm / s, and point Q moves at 2cm / s At the same time, what is the distance from the starting point of PQ B to the starting point B of PQ, respectively, if the starting point of PQ B is equal to the starting point of B?
- 14. In the triangle ABC, AC = 6cm, BC = 8cm, ab = 10cm, D, e, f are the midpoint of AB, BC and Ca, respectively. Calculate the area of triangle def
- 15. As shown in the figure, in △ ABC, ab = 6cm, BC = 8cm, ∠ B = 90 °, point P starts from point a and moves along side AB at a speed of 1cm / s, and point Q starts from point B along the edge of BC Point C moves at a speed of 2 cm / s 1) If the points P and Q start from a and B at the same time, after a few seconds, the area of △ PBQ is equal to 8cm? 2) If points P and Q start from a and B at the same time, and P reaches point B and then continues to move on edge BC, and Q reaches point C and then continues to move on edge Ca, after a few seconds, the area of △ PCQ is equal to 9cm?
- 16. As shown in the following figure, in Δ ABC, ∠ B = 90 ° and point P moves from point a to point B at the speed of 1cm / s along the edge of AB, and point Q starts from point B and moves along the edge of BC to point C Cm / sec.) (1) If P and Q start from a and B at the same time, and P to B continue to move on the BC side, and Q to C continue to move on the CA side. After a few seconds, the area of Δ PCQ is equal to 12.6 cm? 2 Point Q starts from point B and moves along the edge of BC towards point C at a speed of 2 cm / s
- 17. As shown in the figure, in △ ABC, ∠ BAC = 90 °, extend Ba to point d so that ad = half AB, points E and F are the midpoint of edge BC and AC respectively. Find DF = AE
- 18. As shown in the figure, △ ABC is an isosceles right triangle, ∠ a = 90 ° and points P and Q are moving points on AB and AC respectively, and satisfy BP = AQ, D is the midpoint of BC (1) It is proved that △ PDQ is an isosceles right triangle; (2) When the point P moves to where, the quadrilateral apdq is square, and explain the reason
- 19. If the ratio of the three inner angles of a triangle is 9:4:5, then the triangle is ()
- 20. In △ ABC, ∠ a = 1 / 3 ∠ B = 1 / 5 ∠ C, calculate the degree of each internal angle of a triangle