As shown in the figure, in △ ABC, points D and E are on AB and AC respectively, de ‖ BC, ∠ CED = ∠ BDC. (1) prove: △ DCE ∽ CBD; (2) if BC = 2CD, s △ ade = 1, find the value of s △ ABC

As shown in the figure, in △ ABC, points D and E are on AB and AC respectively, de ‖ BC, ∠ CED = ∠ BDC. (1) prove: △ DCE ∽ CBD; (2) if BC = 2CD, s △ ade = 1, find the value of s △ ABC

(1) It is proved that: ∵ de ∥ BC, ∵ EDC = ∥ DCB, ∵ CED = ∥ BDC, ∥ DCE ∥ CBD, ∥ DECD = cdbc, ∥ BC = 2CD, ∥ DECD = 12, ∥ de = 12CD, ∥ DEBC = 12cd2cd = 14, ∥ de ∥ BC, ∥ ade ∥ ABC, ∥ s ∥ ADEs ∥ ABC = de2bc2 = 116, ∥ s ∥ ade = 1, ∥ s ∥ ABC = 16

As shown in the figure, in △ ABC, points D and E are on AB and AC respectively, de ‖ BC, ∠ CED = ∠ BDC. (1) prove: △ DCE ∽ CBD; (2) if BC = 2CD, s △ ade = 1, find the value of s △ ABC

(1) It is proved that: ∵ de ∥ BC, ∵ EDC = ∥ DCB, ∵ CED = ∥ BDC, ∥ DCE ∥ CBD, ∥ DECD = cdbc, ∥ BC = 2CD, ∥ DECD = 12, ∥ de = 12CD, ∥ DEBC = 12cd2cd = 14, ∥ de ∥ BC, ∥ ade ∥ ABC, ∥ s ∥ ADEs ∥ ABC = D

As shown in the figure, in the triangle ABC, D and E are the midpoint of AB and AC respectively, f is the point on BC, EF = 1 / 2BC

∵ D is the midpoint of AB and E is the midpoint of AC
The De is the median of △ ABC
∴DE＝1/2BC,DE∥BC
∴∠DEF＝∠EFC
∵∠EFC＝35
∴∠DEF＝35
∵EF＝1/2BC
∴EF＝DE
∴∠EDF＝∠EFD

There is a pool. The water surface is a square with a side length of 10 feet. In the center of the pool, there is a new reed, which is 1 foot higher than the water surface. If the reed is pulled vertically to the bank, its top just reaches the water surface of the bank. What is the depth of the pool and the length of the reed?

Let the depth of the pool be x feet, then the length of the reed is (x + 1) feet. According to the meaning of the title, we get: (102) & nbsp; 2 + x2 = (x + 1) 2, and the solution is: x = 12, so the length of the reed is: 12 + 1 = 13 (feet). A: the depth of the pool is 12 feet, and the length of the reed is 13 feet

As shown in the figure, we know that the isosceles triangle ADC, ad = AC, B is a point on the line DC, connecting AB, and ab = dB. (1) if the perimeter of △ ABC is 15cm, and ABAC = 23, find the length of AC; (2) if abdc = 13, find the value of Tanc

(1) ∵ ad = AC, ∵ d = ∠ C. ∵ AB = dB, ∵ d = ∠ DAB. ∵ DAB = ∠ d = ∠ C. (1 point) and ∵ d = ∠ D, ∵ DAB ∵ DCA. (1 point) ∵ ADDC = ABAC = 23. (1 point) ∵ 3aD = 2dc. That is, 3aC = 2dc. The perimeter of ∵ ABC is 15cm, that is ab + BC + AC = 15cm, and DB + BC + AC = 15cm. (1 point) ∵ AC = 6cm. (1 point) (2) ∵ abdc = 13, ab = dB That is, BC = 2Ab, (1 point) and DC = 3AB, from (1) △ DAB ∽ DCA, ∽ ABAC = ADDC. ∽ ac2 = 3ab2. (1 point) from BC = 2Ab, BC2 = 4ab2. ∽ AB2 + ac2 = BC2. ∽ ABC is a right triangle. (1 point) and ∠ BAC = 90 °. Tanc = ABAC = 33. (1 point)

As shown in the figure, it is known that AD / / BC AB = DC, and the verification angle B = angle C
Figure A-D
/ \
/ \
B-C is like this

Certification:
Let de ‖ AB intersect BC at point E
Then the quadrilateral abed is a parallelogram
∴∠B=∠DEC,AB=DE
∵AB=CD
∴DC=DE
∴∠DEC=∠C
∴∠B=∠C

It is known that: as shown in the figure, ab ⊥ BC, ad ⊥ DC, the vertical feet are B and D respectively, ∠ 1 = ∠ 2

It is proved that: ∵ ab ⊥ BC, ad ⊥ DC, ∵ B = ≌ d = 90 °, ∵ in △ ABC and △ ADC, ∵ 1 = ≌ 2 ≌ B = ≌ DAC = AC ≌ ABC ≌ ADC (AAS), ∵ AB = ad

As shown in the figure, AB is perpendicular to BC, ad is perpendicular to DC, the perpendicular feet are B, D, angle 1 is equal to angle 2, prove that AB is equal to AD

No picture

As shown in the figure, B is the midpoint of CE, ad = BC, ab = DC. De intersects AB at F

It is proved that: (1) ∵ ad = BC, ab = DC, ∵ quadrilateral ABCD is parallelogram, ∵ ad ∥ BC. (2) ∵ B is the midpoint of CE, ∵ be = BC, ∵ ad = BC, ∥ ad = be, and ∥ ad ∥ BC, ∥ a = ∥ Abe, ∵ in △ ADF and △ bef,