A geometric math problem. Mathematical geometry problems

First make 2 angular bisector lines to intersect at o, and then cross their intersection points to the 2 sides to make a high.

You've made three highs, and then using the knowledge of congruence, you can come up with the three perpendiculars being equal.

Finally, the o and the third point are connected, and the congruence judgment can be used.

2 triangles are given congruence.

It is concluded that 2 angles are equal.

Then the three angular bisectors of the triangle row intersect at one point.

Proof that the point o is made as the height of the three sides: og, oh, oj because the distance from the angle bisector to both sides is equal.

Easy to get og=oh=oj

Then go to the point on the bisector of the hanging angle with equal distances on both sides of the corner.

O is the point (1) on the bisector of the b angle

And point b is also point (2) on the bisector of the angle b

From (1) and (2), the ray bo is the angular bisector of abc.

i.e. the three angular bisectors of ABC intersect at one point.

That is. <>

Proof: It is general to make the bisector of any two angles intersect with a point (there will always be an intersection point between two straight lines that are not parallel in the plane), and the bisector of angle A and angle B is made respectively to cross BC to D, Be to cross AC to be to O, and connect CO to AB to F (this step is very important, pay attention to the connection, the following mainly proves that CO is an angle bisector). The crossing point o is made as a three-sided perpendicular line as og bc in g, oh ac in h, oi ab in i.

abo= cbo, bio= bgo=90, bo=bo, boi bog (corner edge theorem).

oi=og is the same as aoi aoh, oi=oe;

cog coe, oh=og (from this derives the theorem: the distance from the points on the bisector of angles to both sides is equal).

OI=OG=OH (Equivalent Substitution).

OGC= OEC=90, OG=OH, OC=OC Coe Cog (edge and corner theorem).

ocb=∠oca

i.e. CO is the angular bisector of ACB.

From the beginning of the diagram, it can be seen that the three bisectors intersect with one o.

Problem solving.

1: Solution:

Connect OA, OB

c=30 aob=2 c=60 [the central angle of the same arc (ab) is twice the circumferential angle] oa=ob

The triangle OBA is an equilateral triangle.

oa=ab=2

That is: the radius is 2

2: Solution: Connect to CD

adc=90

The triangle ABC is a right triangle, AC=3, BC=4, AB=AC+BC=3, +4=25, AB=5

a=∠a△acd∽△abc

ad：ac=ac：ab

i.e.: ad:3=3:5

ad=

Draw yourself as I say.

Through the point of P to do PA, PB, that is, the two lines of 45°, through P to do the surface of the perpendicular line PO, connect OA, OB, let Pa=PB=X, then the triangle obtained by BPA=60° BPA is an equilateral triangle. ab=x, consisting of the triangle poa and the triangle pob, oa=ob=x 2.

Going back to the triangle OAB, it is clear that OA 2 + OB 2 = AB 2, so the projective angle BOA is 90°

: On the surface of the earth, from place A (45°N, 120°E) to B(45°N, 150°W).

The radius of the latitude circle of the corresponding points of A and B is r=rcos45°=root number 2 2 r The longitude difference is 360°-(120°+150°)=90°, so ab=root2r=root2r=root2(root2 2 r)=r AOB is an equilateral triangle, and the spherical angle is AOB = pie 3 The spherical distance between A and B is pie r 3

These two places are on a circle at 45 degrees north latitude, east longitude 120 west longitude 150 is on the circle inferior arc 90 reading, the diameter of the circle at 45 north latitude is the root number 2 times r, then the distance between the two earth planes is the root number 2 times r divided by 4.

The angle between two places on a circle of radius (root number 2) 2r is 2.

So the distance between two places is 1 4 circumference, and the circumference l=(root number 2) r, so the distance between two places is s = ((root number 2) r) 4

The circumference is more than 8 because of ed bc and ab dc

So ABCD is a parallelogram.

So EB is equal to DC is equal to 4 and ED is equal to BC

8 less, because the quadrilateral cDEB is a parallelogram, so eb=cd=4, and de=bc, so the triangle ade is 2 times less than the circumference of the trapezoid.

Square area = 2r 2

Circular area = r 2 = 22 7r 2

3. The radius of the circle r 2 is 1cm, 2cm, 4cm respectively, and the area of the shaded part is 22 7*(1+2+4)-2*(1+2+4)=22-14=8cm 2

Solution: Let the radius of the three circles from the outside to the inside be R1, R2, and R3 respectively, and the three squares become longer from the outside to the inside, respectively, A1, A2, and A3

Because r3 = 1, a3 = 2

a2=2r2=√2

a1=2√2

r1 = 2 area of the shaded part.

Glad to answer for you :

Analysis: Find the radius of each circle in turn, subtract the area of the corresponding square, and then add it to get the answer.

Answer: 1) -root number 2) + root number 2) -4 + 2 -8 = 8