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Known: as shown in the figure, find a point P on the straight line Mn, so that the distance between the point P and the two sides of ∠ AOB is equal (the method is required to be written, the drawing trace shall be retained, and the conclusion shall be written)
Make the bisector of AOB, and the bisector of AOB intersects the straight line Mn at a point,
As shown in the figure
Drawing with ruler and gauge, as shown in the figure, point P is a point in angle AOB, and straight line Mn parallel OA is drawn through P
Methods: 1. Continuous Op;
2. Take o as the center of the circle and OP as the radius to make the arc intersection OA at point C;
3. Take P and C as the center of the circle and OP as the radius to intersect at point D;
4. Make a straight line Mn through points P and D, then Mn is the result
Proof: (omitted)
As shown in the figure, point P is in ∩ AOB, and points m and N are symmetric points of point P on OA and ob respectively. They connect Mn and intersect OA and ob at points E and f respectively If Mn = 20cm, find the circumference of △ PEF
Connect PE and PF, ∵ m, P are symmetrical about OA, ᙽ EM = EP, the same way: FN = FP, ᙽ perimeter of PEF = PE + EF + pf = me + EF + FN = Mn = 20 cm
As shown in the figure, the angle between the two plane mirrors OA and ob is 110 ° and the light is reflected to the plane mirror ob through the plane mirror OA, and then reflected out, where angle 1 equals angle 2, then the degree of angle 1 is a process of () °
35 degree process ∵ incident angle = reflection angle
∵ 1 = ∵ 2
∴∠1＋∠2＋110°＝180°
∴∠1＝∠2＝35°
As shown in the figure, the reflected light from Mn light source s of plane mirror passes through point a, and all the light passing through point a is drawn S. A.M -------------------------------------------------------------N S. A. M
(1) Take Mn as the axis, make the symmetry point B of point a, connect SA with Mn and connect Ca, then CA is the reflected light after the light from s reaches point C
(2) The straight line passing through a is drawn directly through s point, which is the light passing through a directly through s point
As shown in the figure, A1 is the image of the luminous point a in the plane mirror Mn. Try to draw the light path diagram of a ray from the luminous point passing through point B after being reflected by the plane mirror. (mark the positions of the light-emitting point a and the incident point o)
The symmetry point A1 'of image point A1 is made by passing through the mirror. The incident point (reflection point) is connected by connecting a1p and the mirror surface at point O. the incident light is obtained by connecting A1 ′ o, and ob is the reflected light
As shown in the figure, s is the luminous point and Mn is a plane mirror. Please draw the light from point s and reflected by the plane mirror and passing through point a
Make the symmetrical point of the luminous point s about the plane mirror, that is, the image point s'. Connect the s' and a points to the plane mirror at point O, draw the reflected light along OA, and connect so to draw the incident light, as shown in the figure
As shown in the figure, s is the luminous point and Mn is a plane mirror. Please draw the light from point s and reflected by the plane mirror and passing through point a
Make the symmetrical point of the luminous point s about the plane mirror, that is, the image point s'. Connect the s' and a points to the plane mirror at point O, draw the reflected light along OA, and connect so to draw the incident light, as shown in the figure
It is known that ad ⊥ ob OC bisection ﹣ AOB P is a point on OC, and a straight line Mn is made through point P. OA ob is at point m and N respectively. It is proved that the distance from point P to AO and ad is equal As above
How to do without any plan
Given ∠ AOB and the line L, find the point P so that the distances from P to OA, OB on both sides of ∠ AOB and from line L are equal As the title, ⊙ o ⊙
Firstly, extend OA, OB on both sides of ∠ AOB and intersect with line L at C, D
Make the bisector of angles C and D through C and D, and the intersection point is the P we require
Method of angular bisector:
Take the vertex as the center of the circle, take the equal line segment on the two sides, and get two intersection points
Take two intersection points as the center of the circle and take a larger radius as the arc to get the intersection point
The line connecting the points of intersection and angle is the bisector
RELATED INFORMATIONS
- 1. In the plane rectangular coordinate system, the image of function y = - 3 / 4x + 6 intersects X axis and Y axis at point a and B respectively, line BC and X axis intersect at C, and point C is the midpoint of line OA Find out whether there is a point m on the straight line AB so that △ BCM is an isosceles triangle. If there is, find out the coordinates of point M
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- 8. Known: as shown in the figure, in the right triangle ABC, the angle c = 90 degrees, the point O is the center of the circle, the length of OA is the radius of the circle, and AC, AB intersect with point D.E, and the angle CBD = angle a.bd is the tangent line of circle O. find: if ad: Ao = 8:5, BC = 2, find the length of BD
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- 15. As shown in Fig
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