Axisymmetric geometry problems are explained by math masters!! 1

135 degrees. I just did.

Look. Rotate the triangle APB 90 degrees counterclockwise (how to make the auxiliary line!) Get three called form a"bp"to connect AA",ap",pp", ac so triangle pbp"is an equilateral right-angled triangle, then the angle pp"b is 45 degrees.

In the triangle ap"a"and triangle apc.

aa"=ac a"p"=ap angle aa"p"= angular pac so triangle ap"a"and triangle apc congruence.

Get the AP"=pc

And because ap:bp:pc=1:2:3, so.

ap=a pc=ap"=3a bp=p"b=2a, so use the Pythagorean theorem to get pp"The square = the square of 8a.

AP squared + pp"square = ap"of the squared.

again because of the angle p"pb = 45 degrees.

So the degree of angular apb is 90 + 45 = 135

Solution: Rotate the triangle APB 90 degrees clockwise around point B, turn the original point A to point C, and set the original point P to point Q.

Then the angle bqc is the angle apb, and let cq=ap=k

Then bq=bp=2k, cp=3k

And the angle pbq = 90 degrees, so pq = 2 times the root of 2 times k[(2 2)k], and the angle bqp = 45 degrees (isosceles right triangle).

And because cq=k, cp=3k, (cq, pq, cp length is exactly the number of pythagorean), the triangle qcp is a right triangle, and the angle cqp = 90 degrees.

So the angle bqc = angle bqp + angle cqp = 45 degrees + 90 degrees = 135 degrees.

So the angle apb = 135 degrees.

No way, draw it yourself)

ab=ac∠b=∠acb

eb=ed b= Extinction edb

acb=∠edb

ef‖acae=be

be=deab=ae+be=ac

ed=df, diff ef=ed+df

The quadrilateral aefc is a flat virtual macro calling quadrilateral.

a=∠f<>

3.The following statement is true:

A. If the points A and B are symmetrical with respect to the straight line Mn, then the line segment AB is vertically bisected by MN and the line segment MN is vertically bisected AB

b. If two figures are symmetrical with respect to a certain straight line, then these two figures must be located on both sides of the straight line, and the two figures can also intersect with the straight line respectively.

c. Two triangles symmetrical with respect to a certain straight line are congruent.

d. Congruent triangles are symmetrical about a certain straight line.

Congruent triangles have nothing to do with symmetry because after one of them is rotated by a certain angle, the congruent triangle is congruent, but not symmetrical with respect to a straight line.

It is impossible to bisect the axis of symmetry with the AB line in A.

It is impossible for two figures in b to be on either side of the axis of symmetry, and there are also two figures that intersect the axis of symmetry.

d Two triangles that are symmetrical about a certain straight line must be congruent, while two triangles that are congruent are not necessarily symmetrical if their angles are both facing the same direction.

A: MN bisects AB vertically

b: These two figures can also intersect with straight lines.

d: Congruence is not necessarily related to linear symmetry.

Question 1: If F is perpendicular to AC and the perpendicular foot is H, then the triangle AFH is similar to the triangle ACB, then FH ha=CB AB=3 6=2, and AC= (3 square + 6 square) = 3 5, so ah=(3 2) 5, FH=(3 4) 5, the area of the triangle AFC = (1 2)*AC*FH=(9 8) 5

1.The triangle afc is obviously congruent with the triangle abc, it is a right triangle instead of an isosceles triangle, and the area is 3*6 2=9

After folding EF in half, there is EB=ED, DF=BF, so that it can be calculated...

This is impossible, ab=8, ad=10, so ed>ad>ab>eb conditions should be ab=10, ad=8, so that it can be calculated.

Connect od, because the point e is symmetrical with the point o with respect to the straight line bc, so oed= doc

Because of BC AO, AOB = Lift OBC = 35° and because the point A and point D are symmetrical with respect to the straight line OB, so Lao Bi AOB = DOB = 35°

So oed= doc 90° 2 30° 30°

If there is a problem with the condition you give, it should be ac=bc, and abc is an isosceles right-angled triangle proof: connect cdBecause D is the midpoint of AB, CD=AD and CD AB=AE, the angle A= DCB=45°

So aed afd

So ade= cdf

So edf= edc+ cdf= edc+ ade=90°, i.e. de df

Solution: Connect cd because ae=cfcd=cd a= dcf, so aed is all equal to dcf, so ade= cdf, and because ade+ edc=90°, cdf+ edc=90°, i.e., edf=90°, so de df.

Connecting the CDX three lines and one gives CD vertical ab ad=1 2ab The middle line on the hypotenuse = half of the write edge, i.e., ad=cd, it can be proved that AED is all equal to DCF, so cdf=ade, so EDF=90, i.e., de df

You try to translate a triangle with a 60-degree angle horizontally, note that the two right-angled triangles with a 60-degree angle before and after the translation are congruent, but obviously not axisymmetric, as shown in the figure.

Do you know the meaning of axisymmetry, the nature of nature? Fold a shape along a straight line if it can coincide with another shape.

Axisymmetry, then let's say that these two figures are symmetrical about this straight line, this straight line is called the axis of symmetry, and the points that coincide after folding are the corresponding points (symmetric points), which are called symmetry points. The characteristics of axisymmetric and axisymmetric graphs are the same, and the distance from the corresponding point to the axis of symmetry is equal.

Axisymmetric figures have the following properties: (1) two axisymmetric figures are congruent; (2) If the two graphs form axisymmetry, then the axis of symmetry is the perpendicular bisector of the line connecting the symmetry points;

A straight line that passes through the midpoint of a line segment and is perpendicular to that line segment is called the perpendicular bisector of the line segment. In this way, the following properties are obtained:

1。If two figures are symmetrical with respect to a line, then the axis of symmetry is the perpendicular bisector of the segments connected to any pair of corresponding points.

2。Similarly, the axis of symmetry of an axisymmetric graph is the perpendicular bisector of the line segments connected by any pair of corresponding points.

3。The points on the vertical bisector of a segment are equal to the distance between the two endpoints of the segment.

4。The axis of symmetry is a collection of points at equal distances to both ends of a line segment.

The action can be drawn by drawing one side of the axis of symmetry to the other.

The congruence of two figures can be obtained by drawing the axis of symmetry.

Axial symmetry in life ** (6 photos).

Extend the application of axisymmetric and the significance of function images.

Quite simply, you put two identical triangles on the table at random, and they must be equal, but not axisymmetric.

The 1st floor is still wrong.

Because f(1+x)=f(2-x), the function has symmetry, and the axis of symmetry x=since the number of roots is odd, there must be a root at the axis of symmetry, so there is a solution of x= and the other 100 roots exist symmetrically with respect to the axis of symmetry.

Applying the symmetry of the root, suppose one solution is x=k, and the other root with respect to the axis of symmetry x= is x=3-t

So the first root and the last root add up to 3 instead, the second root and the penultimate root add up to get 3, and so on.

Because f(1+x)=f(2-x), the function has symmetry, and the axis of symmetry x=

So the first root and the last root add up to 3, the second root and the penultimate root add up to 3, and so on.

So get 3*50+