The last question on page 136 of Tianfu mathematics

Make an auxiliary line parallel to ab through P and intersect ad at n, so the parallel angle is ∠ 1 = ∠ NPD, that is ∠ 1 + ∠ 2 = ∠ NPD + ∠ 2 = ∠ B (it is still a parallel angle)

How to write the sixth grade winter vacation homework of Beijing Normal University Edition

You are so stupid
1. Because So
2. Use as long as Just
3. Why It's because (reverse two sentences)

A piece of alloy contains copper and zinc. The mass ratio of copper to zinc is 2:3. Now when 6 g zinc is added, the new alloy is 36 g. the mass ratio of copper to zinc in the new alloy should be determined

(1) The total weight of the original alloy: 36-6 = 30 (g), the weight of copper: 30 × 22 + 3 = 12 (g), the weight of zinc: 30-12 = 18 (g). (2) the weight of copper in the new alloy remains unchanged at 12 g, the weight of zinc is 18 + 6 = 24 (g); the ratio of copper to zinc is 12:24 = 1:2; a: the weight ratio of copper to zinc in the new alloy is 1:2

The last question in 56 aspects of mathematics winter vacation assignment (Beijing Normal University Edition)
The following is a figure composed of three squares with side lengths of 6, 8 and 4 cm respectively to calculate the area of the shadow part

6x6=36 8x8=64 4x4=16

The vertices of the triangle ABC are a (1, - 1,2), B (5, - 6,2), C (1,3, - 1), respectively

The ABC vertices of triangle are a (1, - 1,2), B (5, - 6,2), C (1,3, - 1) AB = √ [(1-5) ^ 2 + (- 1 + 6) ^ 2 + (2-2) ^ 2] = √ 41ac = √ [(1-1) ^ 2 + (- 1-3) ^ 2 + (2 + 1) ^ 2] = 5BC = √ [(5-1) ^ 2 + (- 6-3) ^ 2 + (2 + 1) ^ 2] = √ 106. The angle BAC is obtuse, and D is on the extension line of ca. let BD = X

Given the vertex a (1, - 1,2) B (5, - 6,2) C (1,3, - 1) of triangular ABC, what is the length of high BD on the edge of AC

The ABC vertices are a (1, - 1,1,2), B (5, - 6,2), and C (1,3, - 1) AB = [(1-5) [(1-5) ^ 2 + (-1-1-1-1-1-1-1-1-1-1-1-1-1-1-3-1-3-2 + (1-1-1-1-1-1-1-1-1-1-3-1-1-1-3-3-3-2 - (2-6-6-6-6-3-3-3-3-3-3-3-3-3-3-3-6-6-3-3-2 - (2-6-6-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-3-2 + 4, BD = 5, that is, the high BD on the AC side is equal to 5

In △ ABC, ∠ ACB is the acute angle. Point D is the moving point on the ray BC, connecting ad, making a square ADEF on the right side of ad with AD as one side
As shown in Fig. 1, in △ ABC, ∠ ACB is the acute angle. Point D is the moving point on the ray BC, connecting ad. with AD as one side, make a square ADEF on the right side of AD
Answer the following questions:
(1) If AB = AC, ∠ BAC = 90 & ordm
① When point D is on line BC (not coincident with point B), as shown in Figure 2, the positional relationship between line CF and BD is, and the quantitative relationship is
② When point D is on the extension line of line BC, as shown in Figure 3, is the conclusion still true? Why?
(2) If ab ≠ AC, ∠ BAC ≠ 90 & ordm;, point d moves on line BC
Try to explore: when △ ABC meets a condition, CF ⊥ BC (except for the coincidence of points c and F)? Draw the corresponding figure and explain the reason
(3) If AC = 4, radical 2, BC = 3, under the condition of (2), let the edge De of square ADEF intersect the segment CF at point P, and find the maximum length of segment CP
The important thing is to ask the third question -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --????????????????????????

1) The results show that: (1) the positional relationship between CF and BD is vertical and the quantitative relationship is equal; (2) when point D is on the extension line of BC, the conclusion of (1) still holds. From the square ADEF, we get ad = AF, ∠ DAF = 90 & ordm;; ∫ BAC = 90 & ordm;, ∫ DAF = ∠ BAC, ∩ DAB = ∠ fac, ab = AC, ≌ DAB ≌ fac, ≌ CF = B

In the triangle ABC, the angle ACB is the acute angle, the point D is the point on the ray BC, connecting ad, and making a square ADEF on the right side of ad with AD as one side
If ab ≠ AC, ∠ BAC ≠ 90 ° point d moves on BC, when △ ABC satisfies a certain condition, CF ⊥ BC (except that C and f coincide)

As shown in the figure, ⊿ ADF isosceles right angle, F, a, C, D are in a circle, ⊥ FCA = ⊥ FDA = 45 & amp; sup2;, ⊥ ACB = 45 & amp; ordm;. When only ⊥ C = 45 & amp; ordm;, every point D & nbsp; on the ray BC. F of the square ADEF has CF ⊥ BC

As shown in figure a, in the triangle ABC, the angle ACB is the acute angle, and D is the moving point on the ray BC, connecting ad, with AD as one side and

(1) When point D is on the extension line of BC, the conclusion of (1) still holds (as shown in Figure 3). From the square ADEF, we can get ad = AF, ∠ DAF = 90 °, ∵∠ BAC = 90 °, and ∠ DAF = ∠ BAC ≠ DAB = ∠ fac, and AB = AC, ｜ △ DAB ≌ △ fac, ｜ CF = BD, ∠ CF = AC, CF = FA, CF = FA, CF = FA, CF = FA, CF = FA, CF = FA, CF = FA, CF = FA, CF = FA, CF = FA, CF = fa, CF = FA, CF = FA, CF = FA, CF = FA, CF

As shown in figure a, in △ ABC, ∠ ACB is the acute angle, and point D is the moving point on the ray BC, connecting ad, with AD as one side and making a square ADEF on the right side of AD
(1) If AB = AC, ∠ BAC = 90 °, ① when point D is on line BC (not coincident with point B), as shown in Figure B, the position relationship between line CF and BD is The quantity relation is (2) when point D is on the extension line of line BC, as shown in Fig. C, is the conclusion still true and why? (2) If ab ≠ AC, ∠ BAC ≠ 90 ° point d moves on the line BC. Try to explore: when △ ABC satisfies what condition, CF ⊥ BC (except that points c and f coincide)? And explain the reason

（1）①CF⊥BD，CF=BD … (2 points) so the answer is: vertical, equal In △ bad and △ CAF, ∵ Ba = Ca ∠ bad = ∠ CAF = AF ≌ bad ≌ △ CaF (SAS) (5 points) ∵ CF = BD, ≌ ACF = ∠ ACB = 45 ° and ≌ BCF = 90 ° CF ⊥ BD & nbsp (7 points) (2) when ∠ ACB = 45 ° CF ⊥ BC can be obtained for the following reasons: (8 points) the vertical line of AC passing through point a intersects the straight line of CB at G & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp (9 points) then ∵ ∠ ACB = 45 °∵ Ag = AC, ∵ AGC = ∵ ACG = 45 °∵ Ag = AC, ad = AF, ∵ GAD = ≌ GAC - ≌ △ CaF (SAS) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp (10 points) ≠ ACF = ∠ AGD = 45 °∈ GCF = ∠ GCA + ∠ ACF = 90 ° (12 points)

The last question on page 136 of Tianfu mathematics