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It is known that in the square ABCD, ∠ man = 45 ° and ∠ man rotate clockwise around point A. both sides of it intersect CB, DC (or their extension) at point m and N, ah ⊥ Mn at point H (1) As shown in Fig. 1, when ∠ man point a is rotated to BM = DN, please write down the quantitative relationship between ah and ab directly______ ; (2) As shown in Fig. 2, when ∠ man rotates around point a to BM ≠ DN, does the quantitative relationship between ah and AB in (1) still hold? If not, please write down the reason, if yes, please prove it; (3) As shown in Fig. 3, given ∠ man = 45 °, ah ⊥ Mn at point h, and MH = 2, NH = 3, find the length of ah
(1) As shown in Fig. 2, extend CB to e so that be = DN. ∵ ABCD is a square,
It is known that in the square ABCD, ∠ man = 45 ° and ∠ man rotate clockwise around point A. both sides of it intersect CB, DC (or their extension) at point m and N, ah ⊥ Mn at point H (1) As shown in Fig. 1, when ∠ man point a is rotated to BM = DN, please write down the quantitative relationship between ah and ab directly______ ; (2) As shown in Fig. 2, when ∠ man rotates around point a to BM ≠ DN, does the quantitative relationship between ah and AB in (1) still hold? If not, please write down the reason, if yes, please prove it; (3) As shown in Fig. 3, given ∠ man = 45 °, ah ⊥ Mn at point h, and MH = 2, NH = 3, find the length of ah
(1) As shown in Fig. 2, extend CB to e so that be = DN. ∵ ABCD is a square,
As shown in the figure, m and N are a point on the extension line of the square ABCD side CB and DC with the side length of 1, and dn-bn = Mn (1). It is proved that the angle man = 45 (2) if DP is vertical to an (2) If DP is perpendicular to an and P, prove PA + PC = root two PD (3). If C is the midpoint of DN, write the length of PC directly
Rotate △ ABM with point a as the axis, turn AB to coincide with AD, and set point m to M1
Then △ ABM turns to △ ADM1 (these two triangles are congruent)
So ∠ Mam1 = ∠ DAB = 90 °
Let's look at △ anm and △ anm1
∵AM=AM1,AN=AN,MN=ND+MB=ND+DM1=NM1
∴△ANM≌△ANM1
∴∠NAM=∠NAM1=45°
In the known square ABCD, the angle man = 45 ° and the angle man rotates clockwise around point a, and its two sides intersect respectively CB.DC (or their extension) to M.N, Ah ⊥ Mn at point h, when the angle man rotates around point a to BM ≠ DN, the relationship between ah and AB, and the reason
Proof: proof: take a point G on the extension line of CB, make BG = DN ? square, ABCD ∵ AB = ad, ∵ ADC ? ADC ? ADC ? ADC ? add = 90 ? BG = DN ? ABG ≌ adn (SAS) ≌ (ADN (SAS) ≌ (SAS) ≌ (SAS) ≌ (SAS) ? (SAS) ? (Ag = an ? Dan = Ag = an, + ∠
AB is the diameter of the semicircle, C is a point on the semicircle, D is the midpoint of arc AC, De is perpendicular to AB and E, and am = Mn is proved M is the intersection of De and AC, and N is the intersection of AC and BD
Proof: because D is the midpoint of arc AC
So arc ad = arc DC
Ad B = 1 arc angle
Angle DAC = 1 / 2 arc DC
So angle B = angle DAC
Because AB is the diameter of a semicircle
So angle ADB = angle ade + angle BDE = 90 degrees
Because De is vertical ab
So the angle DEB is 90 degrees
Because angle DEB + angle B + angle BDE = 180 degrees
So angle DAE = angle B
So angle dam = angle MDA
So DM = am
Because BDA = angle MDA + angle MDN = 90 degrees
Angle dam + angle MND = 90 degrees
So angle MDN = angle MND
So DM = Mn
So am = Mn
As shown in the figure, a, B and C are three villages on a highway. The distance between a and B is 100km and that between a and C is 40km. Now a station is built between a and B, and the distance between P and C is XKM (1) The sum of the distance from the station to the three villages is represented by an algebraic expression containing X; (2) If the total distance is 102 km, where should the station be located? (3) To minimize the total distance from the station to the three villages, where should the station be located? What is the minimum?
（1） The total distance is PA+PC+PB=40+x+100- (40+x) +x= (100+x) km;
(2) 100 + x = 102, x = 2, the station is 2km on both sides of C;
(3) When the sum of station and distance is 100 km, C = 100 km
As shown in the figure, a, B and C are three villages on a highway. The distance between a and B is 100km. Now, a station P is set between a and B, and P, B are set up The distance between C is XKM. (1) the sum of the distance from the station to the three villages is expressed by the algebraic formula containing X. (2) if the sum of the distance is 102 km, where should the station be located? (3) to minimize the total distance from the station to the three villages, where should the station be located?
(1) The total distance from the station to the three villages is: (100 + x) (km)
(2) According to the meaning of the question: 100 + x = 102, x = 2
So point P is located 2 km away from the left (or right) side of village C
(3) Because the sum of distances from station p to three villages a, B, C is 100 + X
To minimize 100 + X, x = 0
So station P should be located in village C
As shown in the figure, a, B and C are three stations on a highway. The distance between a and B is 100km, and the distance between a and C is 40 km process
140 or 60
As shown in the figure, a, B and C are three villages on a highway
P. The distance between C and C is x km, PA + Pb = 100 km, PC = x 0 d (1) (1) x + X + 40 + 60-x = x + 100 a: the station should be located at C 2km away from the village, x0d (3) the station should be located at the village C, x0d (2) 100 + x = 102, x = 2 / x0d A: P at the place from C 2km (42km or 38km from point a) x0d (3) because
As shown in the figure, a, B and C are three villages on a highway. The distance between a and B is 100km and that between a and C is 40km. Now a station is built between a and B, and the distance between P and C is XKM (1) The sum of the distance from the station to the three villages is represented by an algebraic expression containing X; (2) If the total distance is 102 km, where should the station be located? (3) To minimize the total distance from the station to the three villages, where should the station be located? What is the minimum?
（1） The total distance is PA+PC+PB=40+x+100- (40+x) +x= (100+x) km;
(2) 100 + x = 102, x = 2, the station is 2km on both sides of C;
(3) When the sum of station and distance is 100 km, C = 100 km
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