As shown in the figure, D is a point on the hypotenuse ab of RT △ ABC, and a circle with the diameter of CD intersects the three sides of △ ABC on e, F and g respectively FG, Fe, connection point (1) It is proved that: ∠ EFG = ∠ B; (2) If AC = 2BC = 4 5, D is the midpoint of AE, and the length of CD is calculated

(1) Proof: connect GD;
∵ CD is the diameter,
∴∠CGD=90°；
∴DG∥BC，
∴∠ADG=∠B；
And ∵ the quadrilateral dgfe is the inscribed quadrilateral of a circle,
∴∠ADG=∠EFG；
∴∠B=∠EFG；
(2) Connect CE, then CE ⊥ AB;
In RT △ ACB, AC = 4
5，BC=2
5；
According to Pythagorean theorem, AB is obtained=
AC2+BC2=10；
Because CE ⊥ AB, according to the projective theorem, AE = ac2 ⊥ AB = 8;
∴AD=DE=4，BE=2；
CE2=AE•BE=16，∴CE=4；
In RT △ CED, CE = 4, de = 4; CD = 4
2．

It is known that in the RT triangle ABC, EF is the median line and CD is the median line on the hypotenuse ab. it is proved that EF = DC

I don't know if you have ever learned a theorem. It's common sense that the central line of the hypotenuse of a right triangle is equal to half of the hypotenuse. If you want to prove it, you should make a rectangle. Its diagonals are equal, and they are equally divided. Therefore, if you take three vertices as diagonal, then the central line is half of the other diagonal line. Therefore, the diagonal of a right triangle is half of the hypotenuse, It must be half of the hypotenuse, so the two lines are equal

As shown in the figure, D is a point on the hypotenuse ab of the right triangle ABC, and the circle with the diameter of CD is called the triangle ABC three sides at e, F, G, connecting EF and FG Verification: angle EFG = angle B

Let's suppose that the circle intersection ab, BC.AC Divided into e, F, G, connected to CE,
∵ arc Ge = arc Ge,  GFE = ∠ GCE,
∵ CD is the diameter,  CED = 90 °,
∴∠A+∠GCE=90°,
∵∠B+∠A=90°,
∴∠B=∠GCE,
That is ∠ GFE = ∠ B

CD is perpendicular to AB and the square of AC is equal to AD

The title should be AC ^ 2 = ad * ab,
The above equation can be changed into AB: AC = AC: AD, and cab is common angle, so triangle CAD and triangle BAC are similar triangles
So angle ACB = angle CDA = 90 degrees
So triangle ABC is a right triangle

In the right triangle ABC, the angle c is equal to 90 ° AC = AB, CD is the center line of AB side, CD is a, what is half of AC

AC = BC, so ∠ a = ∠ B = 45 degrees, because ad = dB, so △ ACD is congruent △ CDB, so ∠ ACD = ∠ BCD = 45 degrees, so CD = ad = a ∠ CDA = 90 degrees, so CA = root sign (a square + a square) = root 2A
So half of AC = root 2A divided by 2 = root 2 times a / 2
I'm so tired. Give me some points

In the right triangle ABC, CD is the hypotenuse AB, the median line CD is 4, AC is 6, then SINB is equal to

The result was 3 / 4

In a right triangle ABC, CD is the center line on the hypotenuse ab. if ∠ B = 30 ° and AC = 5cm, what are the values of AB and CD respectively

Sin30 * AB = AC. so AB = 10, because D is the midpoint of AB, so ad = 1 / 2Ab = 5

In △ ABC, ∠ C = 90 ° De is the perpendicular of AB, ab = 2Ac, and BC = 18cm, then the length of be is______ .

Connect AE,
In ∵ ABC, ∵ C = 90 ° and ab = 2Ac,
∴∠B=30°，∠BAC=60°，
∵ De is the vertical bisector of ab,
∴AE=BE，∠EAD=∠B=30°，AD=BD=1
2AB，
∴△BED≌△AED，
∵∠BAC=60°，∠EAD=30°，
∴∠CAE=∠EAD=30°，
∵AB=2AC，AD=BD=1
2AB，
∴AC=AD，
∴△BED≌△AED≌△AEC，∠B=30°，
∴EC=DE=1
2BE，BC=BE+EC=BE+1
2BE，
Three
2BE=18cm，
∴BE=12cm．
So the answer is: 12cm

As shown in the figure, in the right triangle ABC, ∠ ACB = 90 ° and AC = BC = 8 Starting from point a, point d moves at a speed of 1.5 cm / s along AC direction, and point e starts from point C and moves at a speed of 2 cm / s along CB direction. If points D and e start from a and C at the same time, the area of △ CDE after several seconds is 10 square cm

Let the time be x, then the area s = 1 / 2 (8-1.5x) 2x, and x = 2 / 3 (31 ^ 0.5-4), where "31 ^ 0.5" is the 31 square root

As shown in the figure, in △ ABC, ∠ ACB = 90 °, BC = 3aC = 4, the vertical bisector De of AB intersects the extension line of BC at point E, then CE is

D is the midpoint of ab,
According to Pythagorean theorem, ab = 5
ABC is similar to EBD, BD = 5 / 2, be = 5 / 2 * 5 / 3 = 25 / 6, EC = eb-bc = 25 / 6-3 = 7 / 6,