Verify that the angular bisector of the triangle intersects at a point.

The distance from the center of the circumscribed circle of the triangle to the three sides is equal, and in the triangle, the distance from the straight line passing through one corner to the two sides of the angle is equal, then the angle line is the angle bisector of the angle, and the center of the circle and the three vertices are connected, then these three are the angle bisector, and they intersect at one point - the center of the circle.

Draw a triangle ABC first, and then draw the intersection of angle A and angle B bisector O, connecting OC. Draw OD perpendicular to AB, OE perpendicular to AC, OF perpendicular to BC.

It is known that: 1. The distance from the point on the bisector of the angle to both sides of the angle is equal.

2. Points with equal distances to both sides of the angle are obtained from 1 on the bisector of the angle (angles less than 180 degrees) by 1

od=of, so of=oe is the bisector of the angle c from 2.

Conclusion: The angular bisector of the triangle intersects at one point.

Typing is so tiring.

Draw two bisectors of the triangle, connect the intersection of the other vertex and the two angular bisectors, and prove that this line is an angular bisector, and you're OK.

It is proved according to the circumferential angle equal to three hundred and six.

Using the Seva theorem and the angle bisector theorem.

Know: In ABC, the angular bisector of A and B intersects at the point O;

Verification: The abc angle bisector intersects at the point O.

Proof that the point O is on the angular bisector of a, and the distance from O to ab is equal to the distance from O to ac;

The same can be said: the distance from O to BC is equal to the distance from O to BA.

According to the equivalent substitution, we can see that the distance from O to ac and O to BC is equal, and ac and bc are sides of c, so the point o is on the angular bisector of c.

o is the point on the bisector of the angles a, b, and c in abc.

The proposition is proven. I think that the proof of the wind madness is incorrect, it is the use of the correctness of the proposition to prove the proposition itself. Because the heart of the triangle is defined by knowing that the bisector of the triangle angle intersects at one point.

Know: In ABC, the angular bisector of A and B intersects at the point O;

Verify the spring change or annihilation early: the abc angle bisector intersects at the point O

Proof that the point O is on the angular bisector of a, and the distance from O to ab is equal to the distance from O to ac;

The same can be said: the distance from O to BC is equal to the distance from O to BA.

According to the equivalent substitution, we can see that the distance from O to ac and O to BC is equal, and ac and bc are sides of c, so the point o is on the angular bisector of c.

o is the point on the bisector of the angles a, b, and c in abc.

The proposition is proven.

The intersection of the three angular bisectors of the triangle at one point is called the heart of the triangle, which is the center of the inscribed circle of the triangle.

This is a theorem that does not need to be proved, and it has nothing to do with the congruent triangle mentioned in the head. ,5,Congruent triangle should be two,bisector intersects at one point,There must be two lines equal,This is the theorem,For example:on=oe,know that point O is on the bisector of a,2,Verification:

The three bisectors of the triangle meet at one point.

Verification: The bisector of the three corners of the triangle intersects at one point.

Let the friend ant abc, bf, ag be the bisector of bac and abc, intersect o, connect co and extend ab to e, do op ab, oq bc, or ac

Because AG and BF are angular divisions.

So og=op=or

Because or ac oq bc or=oq oc=oc roc qoc

So rco= qco

The prepared crack to CE is ACB angle division imitation closed line.

Prove it by counter-evidence. Assuming that the third angular bisector is not the focal point, it can be seen from the definition of the angular bisector that the sum of the inner angles of the triangle formed by one side of the triangle and the two angular bisectors is not 180 degrees, and the assumption is wrong. Therefore, the angular draft of each age of any triangle is crossed at one point.

The perpendicular lines on the three sides of the ΔABC are the angular bisectors of ΔABC by crossing the point O, OR, ON bn, and CM.

om=or=on

The point o is on the angular bisector of the bac.

How to verify this, you just draw three diagrams, just draw the bisector of the right-angled, obtuse, and acute triangles, and use a ruler to make the diagram.

Let the angular bisector of any two angles (A and B) intersect at point O, then, according to OA, is the angular bisector of angle A, so that point O is equal to the distance between the two sides of the triangle AB and AC;

In the same way, according to ob is the angle bisector of the angle b, the distance between the o point and the two sides of the triangle ba and bc is equal;

If you connect Co, then since Point O is equal to the distance between the two sides of Ca and Cb of the triangle, Co is the angular bisector of angle C.

Proof is complete.