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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
Method 1
According to the formula of equal proportion, it is concluded that:
AB:GE = （CE+5）:CE
CE = 25/3
BC = 40/3
Area of shadow part = triangular def area - triangular GEC area
=Triangle ABC area - Triangle GEC area
=1/2[(BC*AB)-(CE*GE)] = 65/2
Method 2
According to the formula of equal proportion, it is concluded that:
AB:GE = （CE+5）:CE
CE = 25/3
BC = 40/3
Area of shadow part = triangular def area - triangular GEC area
=Triangle ABC area - Triangle GEC area
=Trapezoidal abeg area
=1/2(AB+GE)*BE
=65/2
As shown in the figure, translate the right triangle ABC along the direction of ray BC to triangle def, and calculate the area of shadow part in the graph!
5 + 8) × 5 △ 2 = 32.5 --- this is the solution of junior high school
The triangle GEC is the part where two triangles coincide. The two triangles subtract the overlapped parts, and the remaining parts are equal, that is, gdfc and abeg are equal. Then only the area of abge is required to know the area of shadow part, that is, (5 + 8 × 5 △ 2 = 32.5)
As shown in the figure, translate the right triangle ABC along the BC direction by the distance be to get the self-called triangle def. Given Ag = 2, be = 4, de = 6, calculate the area of the shadow part
As shown in the figure, from the nature of translation, de = AB = 6, CF = be = 4, ∠ Dec = ∠ B = 90 °, BG = Ab-Ag = 6-2 = 4cm ∵ AC
If AB = 6, BC = 8, be = 4, DH = 3, calculate the shadow area in the graph
First calculate the area of the triangle. Because the triangle ABC is a right triangle, AB * BC = 6 * 8 / 2 = 24. Because it is a translation, ab = De, AC = DF, BC= EF.BC -Be = EC = 8-4 = 4, de-dh = he = 6-3 = 3.3 * 4,2 = 6, because the area of triangular DEF is 24,24-6 = 18, so the area of triangular HEC is 18
In the triangle ABC, D is the midpoint of BC and E is the midpoint of AD. passing through point a is the parallel line of BC, and the extension line of be is intersected with point F and connected with CF. (1) it is proved that AF is equal to DC
Because AF is parallel to BC, triangle AEF is similar to triangle bed, and AF / BD is equal to AE / ed
Therefore, DC / BC = 1
In the triangle ABC, D is a point on the edge of BC, e is the midpoint of AD. through point a, make a parallel line crossing the extension line of CE to F, cut AF = BD, connect BF, and prove that D is the midpoint of BC. If AB = AC, try to judge the shape of quadrilateral afbd
Since AF is parallel to BC and E is the midpoint of AD, triangle AFE is equal to triangle DCE
So AF = CD, so BD = CD, so D is the midpoint of BC
If AB = AC, afbd is a rectangle
In the triangle ABC, D is a point on BC, e is the midpoint of AD. through point a, make a parallel line of BC, intersect the extension line of CE at f and AF = BD, connect BF to prove that D is the midpoint of FC. If AB = AC, try to judge the shape of quadrilateral afbd to prove your conclusion
It is proved that AF / / = BD, so afbd is a parallelogram
Because: AE = ed, ∠ AFC = ∠ BCF, ∠ AEF = ∠ Dec
So: △ AEF ≌ △ Dec
It can be seen that CD = AF = BD
So: D is the midpoint on the BC side
If AB = AC and BD = CD
So: ad ⊥ BC
It can be concluded that the quadrilateral afbd is a rectangle
In △ ABC, AB• AC＝1， AB• BC＝−3． (1) Find the length of AB edge; (2) Find sin (a − b) The value of sinc
（1）∵
AB•
AC=
AB•(
AB+
BC)
=
AB•
AB+
AB•
BC=
AB2-3=1．
∴|
Ab| = 2. That is, the length of AB side is 2. (5 points)
(2) It is known that (1) 2bcosa = 1, 2acos (π - b) = - 3,
acosb = 3bcosa (8 points)
According to the sine theorem, sinacosb = 3sinbcosa (10 points)
∴sin(A-B)
sinC=sin(A-B)
sin(A+B)=sinAcosB-cosAsinB
sinAcosB+cosAsinB=1
2 (12 points)
In the triangle ABC, the vector AB points multiplication vector AC = 1, the vector AB dot multiplication vector BC = - 3 (1) calculates the length of AB edge (2) calculates the value of sin (a-b) / sinc 10 points for answering within one hour
1) Vector AB (vector ac-bc) = 4
Vector AB * vector AB = 4
AB=2
2)AB*AC*cosA=1
AB*BC*cosB=-3
Compare the two forms
(AC*cosA)/(BC*cosB)=-1/3
According to the sine theorem
（sinB*cosA）/(sinA*cosB)=-1/3
-3sinBcosA=sinAcosB
Sin (a-b) / sinc = sin (a-b) / sin (a + b) = 2
In the triangle ABC, ab = AC, BD is perpendicular to D. please specify that the angle CBD = 1 / 2 angle A
Let the midpoint of BC be e
BE=EC AE=AE AB=AC
△ABE≌△ACE
∠AEB=∠AEC=90°
∠EAC+∠C=90°=∠CBD+∠C
So ∠ EAC = ∠ CBD = ∠ EAB = 1 / 2 ∠ a
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