When t=, the quadrilateral is a parallelogram; six
When t=, the quadrilateral is an isosceles trapezoid. eight
2. As shown in Figure 2, if the side length of the square ABCD is 4, the point M is on the side DC, DM= 1, and n is any point on the diagonal AC, then the minimum value of DN+MN is 5.
3. As shown in the figure, in the middle, the point,, is the midpoint. The straight line passing through this point starts from the coincident position, rotates counterclockwise around this point, and intersects this point. The intersection point is the intersection line of this point, and the rotation angle of the straight line is set to.
(1)① When the degrees are equal, the quadrilateral is an isosceles trapezoid, and the length at this time is;
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(standby diagram)
(2) When the degree is 0, the quadrilateral is a right-angled trapezoid, and the length at this time is;
(2) If yes, judge whether the quadrangle is a diamond and explain the reasons.
Solution: (1) ① 30,1; ②60, 1.5;
(2) When ∠ α = 900, the quadrilateral EDBC is a diamond.
∫≈α=∠ACB = 900, ∴BC//ED.∵CE//AB, ∴ quadrilateral EDBC is a parallelogram.
In Rt△ABC, ∠ACB=900, ∠ B = 600, BC = 2, ∠ A = 300.
∴AB=4,AC=2.∴AO= =。 At Rt△AOD, ∴ A = 300, ∴AD=2.
∴BD=2.∴BD=BC. And ∵ quadrilateral EDBC are parallelograms.
∴ Quadrilateral EDBC is a diamond.
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Figure 3
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Figure 2
4. At △ABC, ∠ ACB = 90, AC=BC, straight line MN passes through point C, AD⊥MN is in D, BE⊥MN is in E 。
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Figure 1
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(1) When the straight line MN rotates around the point C to the position of Figure 1, it is proved that: ①△ADC?△CEB；; ; ②DE = AD+BE；
(2) When the straight line MN rotates around the point C to the position in Figure 2, it is proved that DE = ad-be;
(3) When the straight line MN rotates around point C to the position in Figure 3, what is the equivalent relationship between DE, AD and BE? Please write this equivalence relation and prove it.
Solution: (1) ①≈ ACD = ∠ ACB = 90 ∴∠ CAD+∠ ACD = 90 ∴∠ BCE+∠ ACD = 90.
∴∠cad=∠bce∶AC = BC ∴△adc≌△ceb
②∵△ADC?△CEB ∴ce=ad,cd=be ∴de=ce+cd=ad+be
(2) ∫ ∠ ADC = ∠ CEB = ∠ ACB = 90 ∴ ∠ ACD = ∠ CBE and ac = bc.
∴△acd≌△cbe ∴ce=ad,cd=be ∴de=ce-cd=ad-be
(3) When MN rotates to the position shown in Figure 3, DE=BE-AD (or AD=BE-DE, BE=AD+DE, etc. )
∫∠ADC =∠CEB =∠ACB = 90∴∠ACD =∠CBE, and ac = bc,
∴△ACD≌△CBE、∴AD=CE,CD=BE、∴DE=CD-CE=BE-AD.
5. In math class, Teacher Zhang raised a question: As shown in figure 1, the quadrilateral ABCD is a square, point E is the midpoint of the side BC, and the parallel line CF between EF and the outer corner of the square is at point F, which proves AE = ef.
After thinking, Xiao Ming showed a correct method to solve the problem: take the midpoint m of AB and connect me, then AM=EC, which is easy to prove, so.
On this basis, the students did further research:
(1) Xiaoying proposed: As shown in Figure 2, if "point E is the midpoint of the BC side" is changed to "point E is any point on the BC side (except B and C)", other conditions remain unchanged, then the conclusion "AE=EF" still holds. Do you think Xiaoying's point of view is correct? If it is correct, write the proof process; If not, please explain the reasons;
(2) Xiao Hua proposed that, as shown in Figure 3, point E is any point on the BC extension line (except point C), and the conclusion of "AE=EF" still holds. Do you think Xiaohua's view is correct? If it is correct, write the proof process; If not, please explain why.
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Figure 1
Solution: (1) is correct.
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Proof: Take a point on the chessboard, make it and connect it.
. , .
Is the bisector of the outer corner.
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Figure 2
, ,
. (ASA)。 .
(2) correct.
Proof: take a point on the extension line of.
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Figure 3
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A quadrilateral is a square.
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(ASA)。
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6. As shown in the figure, on ray MB, MB=9, A is a point outside ray MB, AB=5, and the distance from A to ray MB is 3. The moving point P moves from M to ray MB at the speed of 1 unit/second, and let the moving time of P be t. 。
Find the t value of (1)△ PAB as an isosceles triangle; (2)△ PAB is the t value of right triangle;
(3) If AB=5 and ∠ ABM = 45, and other conditions remain unchanged, directly write the t value of △ PAB as a right triangle.
7. isosceles trapezoid ABCD, in AD‖BC, E is the midpoint of AB, the intersection point E is EF‖BC, and CD intersects at point F, AB = 4, BC = 6, ∠ B = 60.
(1) Find the distance from point E to BC; (2) Point P is a moving point on line segment EF, if it intersects with P, it is PM⊥EF, BC is at point M, if it intersects with M, it is MN‖AB, ADC is at point N, connected with PN, and EP = X.
① Does the shape of △PMN change when the N point is on the line segment AD? If it is unchanged, find the perimeter of △PMN; If yes, please explain why.
② When n points are on the line DC, is there a p point to make △PMN an isosceles triangle? If it exists, request all values of x that meet the requirements; If it does not exist, please explain why.
① ② 1
① ② 1 2
3 2 3
8. As shown in the figure, cm, cm and point are known as the midpoint of.
(1) If point P moves from point B to point C at a speed of 3cm/s on BC line and point Q moves from point C to point A on CA line.
(1) If the moving speed of point Q is equal to the moving speed of point P, is it the same after 1 s, please explain the reason;
② If the moving speed of point Q is not equal to that of point P, when the moving speed of point Q is what, can it be congruent?
(2) If point Q starts from point C at the speed of ②, and point P starts from point B at the same time at the original speed, and both move counterclockwise along three sides, how long will it take for point P to meet point Q for the first time?
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Solution: (1) ①÷seconds, ∴ cm.
∵ cm, the midpoint of the point, ∴ cm.
* cm, ∴. ∴ cm
Here we go again: ∴. ∴
②:, ∴ and:, and then,
∴ Point, the time seconds of point movement, ∴ cm/sec.
(2) Suppose the points meet for the first time after seconds, and get seconds from meaning, meaning and solution.
∴ The point * * * moved by centimeters. ∵, ∴ a little bit, met on the side,
Seconds later, the point meets the point at the edge for the first time.
7. As shown in figure 1, in the isosceles trapezoid,, is the midpoint and the intersection point is the intersection point. ,, Find: (1) Find the distance from point to point;
(2) A point is a moving point on a line segment, passing through it, crossing the line at that point, connecting and setting.
① When the point is on the line segment (as shown in Figure 2), does the shape of the point change? If it does not change, the calculated perimeter; If yes, please explain the reasons;
② When the point is on the line segment (as shown in Figure 3), is there a point that makes it an isosceles triangle? If it exists, request all the values that meet the requirements; If it does not exist, please explain why.
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Figure 4 (Standby State)
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Figure 5 (Standby State)
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Figure 1
Figure 2
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Figure 3
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(Question 25)
The solution (1) is shown in figure 1, and the passing point is at the midpoint of point \ u.
In the middle.
Figure 1
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In other words, the distance from point to point is
(2)① When a point moves on a line segment, its shape remains unchanged.
∵ ∴
In the same way.
As shown in fig. 2, this point has ended.
Figure 2
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∴ ∴
Then.
Yes,
The circumference of ∴ =
② When a point moves on a line segment, its shape will change, but it is still an equilateral triangle.
As shown in Figure 3, when it is completed in, then
Just like ①, ∴∵ is an equilateral triangle, ∴
At this moment,
Figure 3
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Figure 4
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Figure 5
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As shown in fig. 4, at this time,
As shown in fig. 5, when again
Therefore, this point coincides with a right triangle.
At this moment,
In summary, when or 4 or, it is an isosceles triangle.