Urgent!!! Math Proof Proves that the distance from the right angle of a right triangle to the midpoi

This can be demonstrated by making an circumscribed circle.

Because the triangle is a right triangle, the hypotenuse of the right triangle.

is the diameter of one of its circumscribed circles.

According to known conditions, the midpoint of the hypotenuse is the center of the circumscribed circle.

Thus, the midpoint of the hypotenuse and the vertex of the right angle are connected.

This line is the radius of this circle,—— which is naturally equal to half the diameter!

That's half the hypotenuse of a right triangle!

That means that the distance from the midpoint of the hypotenuse of a right triangle to the three vertices is equal!

Let a(a,0),b(0,b) ab be the midpoint m, then.

The coordinates of point m are <>

Swift lifting. <>

So the mu bureau <>

The circumference of a circle diameter is 90°, the center of the circle is the midpoint of the hypotenuse, and the midpoint of the right angle to the hypotenuse is the radius diameter is twice the radius, so the distance from the right angle of the right triangle to the midpoint of the hypotenuse is half of the hypotenuse.

There are many ways to do this, and here are 4 for reference:

1. The midpoint of a right triangle is exactly its outer center, that is, the center of the circumscribed circle, and the distance from the center of the circle to the perimeter of each point is equal, so the distance to the three vertices is equal.

2. If you are not afraid of trouble, you can use the cosine theorem to calculate it, and you can get the same result.

3. Connect the right-angled fixed point and the hypotenuse midpoint and extend it to equal lengths, and the same result can be proved with congruent triangles.

4. The key coordinates of the hypotenuse can be obtained by taking the right-angled edge as the coordinate axis and establishing the coordinate system, and the results can be obtained through the distance formula between the two points. ......

Through the midpoint of AB (named D) as DH perpendicular to BC, then DHB=90·, and C=90·, so DH is parallel to AC, because D is the midpoint of AB, so DH is the median line of δACB, then H is the midpoint of CB, because DH is perpendicular to CB, so DH is the perpendicular bisector of CB, so; The perpendicular bisector of the BC side intersects at the midpoint of the hypotenuse AB.

Draw triangle ABC, angle C 90

As the BC perpendicular bisector EF, cross BC to F and AB to E because AC is perpendicular to BC and EF is perpendicular to BC

So ac is parallel to ef, and because f is the midpoint of bc, e is the midpoint of ab, and e is e, and eg is perpendicular to g

Obviously, g is the midpoint of ac, so eg is the perpendicular bisector of ac, so the perpendicular bisector of the two right-angled sides of a right-angled triangle intersects at the midpoint of the hypotenuse.

Using the cosine function in the trigonometric function to calculate, cosa is equal to the adjacent edge of the angle a than the hypotenuse, so cosa = 350 1053 =, check the trigonometric function value table, you can get that the angle a is equal to the degree.

with trigonometric functions.

Assuming the angle is a, then cosa = 350 1050

a is approximately equal to degrees.

Another angle can be found using the sinusoidal theorem.

c/sin90=b/sinx，sinx=c/a=，x=

90-x=

Sine theorem: c sinc=b sinb is 1053 1=350 sinb b 19° and the other angle is 90-19=71°