![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in the figure, in RT △ ABC, ∠ C = 90 °, BC = 4, AC = 4, now translate △ ABC along CB direction to △ a ′ B ′ C ', if the translation distance is 3 (1) Calculate the area of the overlapped part of △ ABC and △ a ′ B ′ C '; (2) If the translation distance is x (0 ≤ x ≤ 4), find the area y of the overlapping part of △ ABC and △ a ′ B ′ C ', then what is the relationship between Y and X
（1）∵∠C=90°，BC=4，AC=4，
ν Δ ABC is an isosceles right triangle,
∴∠ABC=45°，
∵ a ′ B ′ C ′ is obtained by △ ABC translation,
∴△ABC≌△A′B′C′，
∴∠C=∠A′C′B′=90°，
∴∠BOC′=45°，
ν Δ BOC ′ is an isosceles right triangle,
∵BC′=BC-CC′=4-3=1，
∴S△BOC′=1
2×1×1=1
2，
That is, s shadow = 1
2；
(2) According to (1), the overlapped part of two triangles is isosceles right triangle,
Then s shadow = 1
2（4-x）2．
In the RT triangle ABC, the angle ACB is 90 degrees, CD is perpendicular to D, e is the midpoint of AC, the extension of ED intersects the extension of CB at P It was proved that the square of PD = Pb × PC
It is proved that ∈ EC is a right angled edge
∴∠ECD+90°=∠EDC+90°,∴∠PCD=∠PDB
And ? P = ∵ P, PCD ∵ PDB
PC: PD = PD: Pb, the square of PD = Pb × PC
In △ ABC, ad is the bisector of ∠ BAC, e and F are points on AB and AC respectively, and ∠ EDF + ∠ EAF = 180 ° and de = DF
It is proved that if D is used as DM ⊥ AB, in M, DN ⊥ AC in n,
That is ∠ EMD = ∠ fnd = 90 °,
∵ ad bisection ∵ BAC, DM ⊥ AB, DN ⊥ AC,
ν DM = DN (angular bisector property),
∵∠EAF+∠EDF=180°，
∴∠MED+∠AFD=360°-180°=180°，
∵∠AFD+∠NFD=180°，
∴∠MED=∠NFD，
In △ EMD and △ fnd
∠MED＝∠DFN
∠DME＝∠DNF
DM＝DN ，
∴△EMD≌△FND（AAS），
∴DE=DF．
In △ ABC, ad is the bisector of ∠ BAC, e and F are points on AB and AC respectively, and ∠ EDF + ∠ EAF = 180 ° and de = DF
It is proved that if D is used as DM ⊥ AB, in M, DN ⊥ AC in n,
That is ∠ EMD = ∠ fnd = 90 °,
∵ ad bisection ∵ BAC, DM ⊥ AB, DN ⊥ AC,
ν DM = DN (angular bisector property),
∵∠EAF+∠EDF=180°，
∴∠MED+∠AFD=360°-180°=180°，
∵∠AFD+∠NFD=180°，
∴∠MED=∠NFD，
In △ EMD and △ fnd
∠MED＝∠DFN
∠DME＝∠DNF
DM＝DN ，
∴△EMD≌△FND（AAS），
∴DE=DF．
It is known that in △ ABC, D, e and F are the midpoint of the three sides, and Ag is the height on the BC side. It is proved that the quadrilateral dgef is isosceles trapezoid
It is proved that ∵ D and F are the midpoint of edge AB and AC respectively,
/ / DF ∥ BC is DF ∥ Ge,
∵DF=BE=1
2BC≠GE，
The quadrilateral dgef is trapezoid,
∵ E and F are the midpoint of AC and BC respectively,
∴EF=1
2AB，
∵ AG is the height on the side of BC,
△ ABG is a right triangle,
∴DG=1
2AB，
∴EF=DG，
The quadrilateral dgef is isosceles trapezoid
As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC edge, E.F is the midpoint of AB and AC respectively, connecting de and DF. Verification: the quadrilateral AEDF is a diamond
It is proved that D, e and F are the midpoint of BC, AB and AC respectively
So we have DF ‖ and = AB / 2
De ‖ and = AC / 2
AE=AB/2
AF=AC/2
So DF ‖ and = AE, de ‖ and = AF
Therefore, by definition, we know that the quadrilateral AEDF is a diamond
I'm tired of typing,
1. As shown in the figure, in the triangle ABC, ad is the bisector of the angle BAC, de / / AC, intersecting AB at e, DF / / AB intersecting AC at F, proving that the quadrilateral AEDF is a diamond
It is easy to know that quadrilateral is parallelogram;
Ad bisects ∠ FAE, so ∠ fad = ∠ DAE, and DF / / AB, so ∠ DAE = ∠ FDA
Therefore, ﹤ FAD= ﹤ FDA, FA=FD;
Therefore, a quadrilateral AEDF is a diamond
It is known that in △ ABC, D, e and F are the midpoint of the three sides, and Ag is the height on the BC side. It is proved that the quadrilateral dgef is isosceles trapezoid
It is proved that ∵ D and F are the midpoint of edge AB and AC respectively,
/ / DF ∥ BC is DF ∥ Ge,
∵DF=BE=1
2BC≠GE，
The quadrilateral dgef is trapezoid,
∵ E and F are the midpoint of AC and BC respectively,
∴EF=1
2AB，
∵ AG is the height on the side of BC,
△ ABG is a right triangle,
∴DG=1
2AB，
∴EF=DG，
The quadrilateral dgef is isosceles trapezoid
It is known that in △ ABC, D, e and F are the midpoint of the three sides, and Ag is the height on the BC side. It is proved that the quadrilateral dgef is isosceles trapezoid
It is proved that ∵ D and F are the midpoint of edge AB and AC respectively,
/ / DF ∥ BC is DF ∥ Ge,
∵DF=BE=1
2BC≠GE，
The quadrilateral dgef is trapezoid,
∵ E and F are the midpoint of AC and BC respectively,
∴EF=1
2AB，
∵ AG is the height on the side of BC,
△ ABG is a right triangle,
∴DG=1
2AB，
∴EF=DG，
The quadrilateral dgef is isosceles trapezoid
It is known that in △ ABC, D, e and F are the midpoint of the three sides, and Ag is the height on the BC side. It is proved that the quadrilateral dgef is isosceles trapezoid
It is proved that ∵ D and F are the midpoint of edge AB and AC respectively,
/ / DF ∥ BC is DF ∥ Ge,
∵DF=BE=1
2BC≠GE，
The quadrilateral dgef is trapezoid,
∵ E and F are the midpoint of AC and BC respectively,
∴EF=1
2AB，
∵ AG is the height on the side of BC,
△ ABG is a right triangle,
∴DG=1
2AB，
∴EF=DG，
The quadrilateral dgef is isosceles trapezoid
RELATED INFORMATIONS
- 1. It is known that ad is an angular bisector of △ ABC, where de ‖ AC intersects AB at point E and DF ‖ AB intersects AC at point F Verification: quadrilateral AEDF is rhombic
- 2. In the graph + equilateral triangle ABC, if D, e and F are on BC, AC and ab respectively, and CD = BF, BD = CE, then what degree is the angle EDF
- 3. AB equals BC equals CD equals de equals EF equals FG equals GA. if angle DAE is a, calculate a value In triangle ade
- 4. In the triangle ABC, D is the midpoint of BC and DM is perpendicular to DN. If BM + CN = DM + DN, verify Ad = 1 / 4 (AB + AC)
- 5. As shown in the figure, in the parallelogram ABCD, the bisector of ∠ bad intersects with edge BC at point E, and bisector of ∠ ABC intersects with edge ad at point F. please prove that the abef of quadrilateral is diamond
- 6. As shown in the figure, it is known that m and N are two points on the edge BC of △ ABC, and BM = Mn = NC is satisfied. A straight line parallel to AC intersects the extension line of AB, am and an at points D, e and f respectively. It is proved that EF = 3DE
- 7. As shown in the figure, △ ABC, ab = AC, points m and N are on the straight line where BC is located, and am = an Excuse me: BM = CN? Please state the reasons
- 8. In △ MNB, BN = 6, points a, C, D are on MB, Nb, Mn respectively, and the quadrilateral ABCD is a parallelogram, and ∠ NDC = ∠ MDA, then the circumference of quadrilateral ABCD is () A. 24 B. 18 C. 16 D. 12
- 9. Known: as shown in the figure, ab ∥ CD, and ab + CD = BC, M is the midpoint of AD Confirmation: BM ⊥ cm
- 10. She is an English girl
- 11. It is known that in △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of AC, AE ⊥ BD is at e, and extended AE is crossed with BC at F. it is proved that ∠ ADB = ∠ CDF
- 12. As shown in the figure, the turning point of a highway is an arc. The point O is the center of the arc, AB is equal to 120m, C is the point on AB, O is the point on AB, the perpendicular foot is D, and CD is equal to 20m. Calculate the radius of the curve
- 13. 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
- 14. As shown in the figure, in the triangle ABC, the vertical bisector L1 of the AB side In the triangle ABC, the vertical bisector L1 of AB side intersects BC at D, L2 of AC side intersects BC at e, Li and L2 intersect at point O. the perimeter of triangle ade is 6 and that of OBC of triangle is 16. Find the length of Ao
- 15. As shown in the figure, in △ ABC, ∠ ACB = 90 °, point E is the midpoint of AB, connecting CE, passing through point E as ed ⊥ BC at point D, taking a point F on the extension line of De to make AF = CE
- 16. In the triangle ABC, the angle c = 90 degrees, D is a point on the edge AC, and ad = 1, angle abd = 45 degrees. Let the angle CBD = a, BC = X. when Tana = 3 / 5, find x =? When x = 0.16, find Tana =?
- 17. As shown in the figure, translate the right triangle ABC along the BC direction to the position of the right triangle def, ab = 8, be = 5, Ge = 5
- 18. As shown in the figure, in △ ABC, ab = AC, D is the midpoint of BC, points E and F are on AB and AC respectively, and AE = AF, it is proved that de = DF
- 19. As shown in the figure △ ABC, ab = AC point D is a point on CB extension line ∠ ADB = 60 ° point E is a point on AD and de = dB, it is proved that AE = be + BC Pictures can be drawn according to the theme!
- 20. As shown in the figure, in the RT triangle ABC, ∠ = 90 ° ad bisect ∠ BAC, and ∠ B = 3 ∠ bad, find the degree of ∠ ADC