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In triangle ABC, AF: FB = BD: DC = Ce: AE = 3:2, be, CF and ad intersect point igh on BC, AC and ab respectively, and the area of IgH is 1
As shown in the figure, the black area in the middle is 1. Do DX, ey and FZ parallel to CF, ad and be respectively. Take the triangle AHC, we can see that ah: HD = AF: FX. And BX: XF = BD: DC = 3:2, so AF: FX = 15:4 (self calculation). That is ah: HD = 15:4, so the area of triangle AHC is 15 / 19 * 1 / 3 = 15 / 57 of ABC
If AB = 3cm, BC = 4cm, AF = 5cm, and ab ⊥ bcaf ⊥ AC quadrilateral cdef is a square, the area of square cdef is ()
A25cm² B50cm² C75m² D100cm²
AC=5 AF=5
CF = 5 radical (2)
The area of cdef was 50
In the RT triangle ABC, the quadrilateral cdef is a square, and the points D, e and F are on BC, AB and AC respectively, and a
In the RT triangle ABC, the square is a quadrilateral cdef with ∠ C = 90 ° and the points D, e and F are on AC, AB and BC respectively, and AE = A and be = B. the sum of the areas of △ ade and △ EFB is calculated
Let ad = a · K ∵ Δ ade ∵ Δ ACB ∵ Δ EFB ∵ DC = B · K, CF = B · K, FB = B ∵ 178; · K / a ∵ a ∵ 178; = (a · K) ∵ 178; + (B · K) ∵ 178;; k = 1 / (a ∵ 178; + B ∵ 178;)] Δ ade + Δ EFB = ad · de / 2 + ef
Known: as shown in the figure, in the triangle ABC, the angle ACB = 90 degrees, D, e, f are respectively the midpoint of AB, AC, BC, prove: the quadrilateral cdef is a rectangle
A triangle is on the lower left B triangle and the lower right C triangle and is the midpoint of the right angle D AC side e AB side f CB side
If you can: the second question is known: as shown in the figure, in ladder ABCD, AD / / BC, EF and Mn are divided vertically and equally, e, F, m and N are the midpoint of AD, BC, BD and AC respectively
Verification: ab = CD
D. E is the midpoint of AB and AC respectively, then De is the median line of triangle, so de / / BC
And de = 1 / 2BC
De is parallel to BC
Then the quadrilateral cdef is a rectangle
As shown in the triangle ABC, the angle ACB = 90 degrees, the intersection of the angle bisector AD and the high CH and the f point, the de perpendicular to AB and E, prove that the quadrilateral cdef is a diamond
CF ‖ De, (DU ⊥ AB) ⊿ ACD ≌ AED (AAS) ≌ CD = DE.AC ＝AE ⊿AEC≌⊿AFE（SAS）
∴CF＝FE.⊿CDF≌⊿EDF（SSS）∴∠FED＝∠FCD＝∠EDB,∴FE‖CD,
∵ CF ∥ De, Fe ∥ CD, ∥ cdef is a parallelogram, and CD = De, cdef is a diamond
Given that two points a (- 2,0) B (4.0) P in the plane rectangular coordinate system are on the straight line y = 1 / 2x + 5 / 2 and the triangle ABP is a right triangle, the point P coordinate can be obtained
There should be three cases: angle a right angle: substituting x = - 2 into y = 1 / 2x + 5 / 2 to get y = 3 / 2 ﹣ having P1 (- 2,1.5) angle b right angle: substituting x = 4 into y = 1 / 2x + 5 / 2 to get y = 9 / 2 ﹣ having P2 (4,4.5) angle P right angle: then p is on a circle with (1,0) as the center and radius of 3 / / have you learned? Diameter corresponds to right angle, have you not learned
As shown in the figure, given point a (- 1,0) and point B (1,2), point P is determined on the coordinate axis so that △ ABP is a right triangle, then point P satisfying such condition has ()
A. 2 b. 4 C. 6 D. 7
① Taking a as the right angle vertex, a straight line through a can be perpendicular to AB and intersect with the coordinate axis at one point, which meets the requirements of point p; taking B as the right angle vertex, a straight line through B can be perpendicular to AB and intersect with the coordinate axis at two points, which also meets the requirements of point p; taking P as the right angle vertex, a circle can be drawn with ab as the diameter, and there are three intersections with the coordinate axis. Therefore, there are six points P meeting the conditions C
When the image of a first-order function passes through the point ＜ - 3,7 ＞ and intersects with the coordinate axis, and forms an isosceles right triangle with the coordinate axis, the analytic expression of the first-order function is obtained
If the triangle is in the second quadrant, let y = x + B,
7=-3+b
b=10
The analytic formula is y = x + 10
If the triangle is in the first quadrant, let y = - x + B
7=3+b
b=4
The analytic formula is y = - x + 4
As shown in the figure, a (- 1,0), B (1,2), determine the point P on the coordinate axis so that the triangle ABP is a right triangle,
There are three types to consider
1) When angle a is a right angle, point P only falls on the negative half axis of Y axis, and the linear equation of AP is as follows:
Y=-[(1+1)/(2-0)]*(X+1)=-(X+1),
When x = 0, y = - 1
Then the point P coordinates are (0, - 1)
2) When angle B is a right angle,
The equation of line Pb is Y-2 = - (x-1)
When y = 0, x = 3
When x = 0, y = 3
Then the coordinates of point P are (3,0) or (0,3)
3) When the angle P is a right angle, there are three cases,
That is, draw a circle with ab as the diameter, intersect one point on the X axis and two points on the Y axis,
At this time, the coordinates of the center of the circle are:
X=(-1+1)/2=0,Y=(0+2)/2=1.
Radius r = √ [(1 + 1) ^ 2 + (2) ^ 2] / 2 = √ 2
Then the circle equation is x ^ 2 + (Y-1) ^ 2 = 2
When x = 0, Y1 = √ 2 + 1, y2 = - √ 2 + 1
When y = 0, x = 1,
Then, the coordinates of point P are (0, √ 2 + 1), (0, √ 2 + 1), (1,0)
Given that a function of degree passes through point P (0, - 2), and the area of a right triangle cut by two coordinate axes is 3, the expression of a function of degree is obtained
Simple,
Area 3 = 1 / 2 * 2 * x calculate x = 3 (which 2 is the intercept of Y axis)
So the intersection of the line and the X axis is (plus or minus 3,0)
A linear function expression is
1.y=(2/3)X-2
2.y=-(2/3)x-2
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, ∠ B = 90 °, ab = 3, AC = 5, fold △ ABC so that point C coincides with point a, and the crease is De, then the perimeter of △ Abe is______ ．
- 2. It is known that: as shown in the figure, point D is a point on the edge AC of △ ABC, passing point D makes de ⊥ AB, DF ⊥ BC, e and F are perpendicular feet, then passing point D makes DG ∥ AB, intersecting BC at point G, and de = DF. (1) prove: DG = BG; (2) prove: BD bisects EF vertically
- 3. RT, in triangle ABC, ab = AC = 10, point D on BC, de parallel AC, DF parallel AB, find the perimeter of quadrilateral AEDF
- 4. As shown in the figure, in △ ABC, de ‖ BC, ad: DB = 3:2 (1) find the value of DEBC; (2) find the value of bced & nbsp; of s △ ADEs quadrilateral
- 5. As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm, now fold △ ABC to make vertex A and B coincide, then the length of crease De is () cm A. 52B. 154C. 158D. 5
- 6. Triangle ABC, AC = BC, angle c = 90 degree, point D on BC, fold triangle ACD along ad, point C just falls on point E on AB, if AB = 6cm Finding the perimeter of triangle deb The picture can't be sent out
- 7. Original proposition: √ a ＞ B, then a ＞ B (write its inverse proposition, no proposition, inverse no proposition)
- 8. If a = B, then a = B (fill in "true" or "false") proposition, its inverse proposition is, it is (fill in "true" or "false") proposition
- 9. If three line segments a, B and C satisfy a square = C square - b square, is the triangle formed by these three line segments a right triangle? Why? Please explain the process and reasons,
- 10. Let a and B be the lengths of two right angles of a right triangle and (A2 + B2) (3 under the root of A2 + b2-4) = - 12, then the value of A2 + B2 is Thank you for your correct answer!
- 11. In the triangle ABC, BD.CD They are the diplomatic bisector of angle ABC and angle ACB
- 12. As shown in the figure, in △ ABC, BD and CD divide ∠ ABC and ∠ ACB equally, and try to explain ∠ d = 90 ° + & # 189; ∠ a
- 13. As shown in the figure, BD is the extension of the intersection of the middle line AE ⊥ BD on the edge AC of △ ABC at the point ECF ⊥ BD. be + BF. = 2bd is proved at the point F It's sloppy
- 14. In ABC, BD is the inner bisector of angle B, CE is the outer bisector of angle c, AF is perpendicular to BD, Ag is perpendicular to CE, FG and ABC?
- 15. (1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, the perpendicular feet are f, G, connecting FG (1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, and the perpendicular feet are f, G, connecting FG. (2) if C
- 16. It is known that in △ ABC, points D, e and F are points on BC, AB and AC respectively, AF is parallel to ED, and AF = ed, extend FD to point G, so that DG = FD, and verify that ED and Ag are equally divided,
- 17. In △ ABC, ∠ C = 90 ° and the intersection line BC of AB is at D. if ∠ bad - ∠ DAC = 22.5 °, then the degree of ∠ B is______ ．
- 18. In △ ABC, the vertical bisectors of AB and AC intersect BC at points E and f respectively. If ∠ BAC = 115 °, then ∠ EAF=______ Degree
- 19. If the inscribed circle of RT triangle is tangent to the hypotenuse AB at D, and ad = 1, BD = 2, then s ABC =? AB is the hypotenuse If the inscribed circle of RT triangle and hypotenuse AB are tangent to D, and ad = 1, BD = 2, then s ABC =? AB is hypotenuse How to use area
- 20. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 80 °, point P is within △ ABC, ∠ PBC = 10 °, PCB = 30 °, calculate the degree of ∠ BAP What are you doing?