It is known that in △ ABC, D is a point on AB, and Da = DB = DC

It is known that in △ ABC, D is a point on AB, and Da = DB = DC

∵ in ᙽ ABC, D is the point above AB, and Da = DB = DC,
∴CD=1
2AB，
The △ ABC is a right triangle

In the right triangle ABC, the angle ACB = 90 degrees, CD is perpendicular to AB and the perpendicular foot is the point D. It is proved that the square of AC = ad * AB and the square of BC = BD * AB (projective theorem)

Triangle ACD is similar to triangle ABC
AD/AC=AC/AB
So the square of AC = ad * ab
Similarly, it can be proved that the square of BC = BD * ab

In the right triangle ABC, CD is perpendicular to AB and the perpendicular foot is point D. It is proved that ab square = ad square + BD square + 2CD square

Using Pythagorean theorem, in RT △ ABC, there is AC square + BC square = AB square,
In RT △ ACD, there is ad square + CD square = AC square
In RT △ BCD, there is BD square + CD square = BC square
So there is ad squared + CD squared + BD squared + CD squared = AC squared + BC squared = AB squared
Then AB square = ad square + BD square + 2CD square

Solving triangles: (1) It is known that in △ ABC, ab = 15cm, AC = 24cm, ∠ a = 60 °. Find the length of BC (2) Given △ ABC, ab = 13, BC = 14, AC = 15, find the high ad on BC edge (3) In △ ABC, CD ⊥ AB is in D. if CD2 = ad · dB, it is proved that △ ABC is a right triangle

(1) As shown in the figure: ∵ a = 60 °, AC = 24cm,  BC = AC · sin60 ° = 24 × 32 = 123; (2)

It is known that in △ ABC, D is a point on AB, and Da = DB = DC

∵ in ᙽ ABC, D is the point above AB, and Da = DB = DC,
∴CD=1
2AB，
The △ ABC is a right triangle

As shown in the figure, △ ABC is an isosceles right triangle, ab = AC, with AC as the edge, make ∠ CAD, make ad = AC, ∠ CAD = 30 ° and connect dB, DC It is shown that: (1) △ abd is an equilateral triangle; (2) BCD is equal to CAD

(1)  bad =  Cab -  CAD = 90 ° - 30 ° = 60 ° and ? ad = AC  abd is an equilateral triangle / * according to: corner edge * / (2) ? ad = AC ? ACD is an isosceles triangle ? ACD = ∠ ADC = (180 ° - 30 °) / 2 = 75 ° ? ACB = ∠ ABC = (180 ° - 90 °) / 2 = 45 °  BCD = ∠ ACD -

The length of the rectangle ABCD is 12 cm and the width is 5 cm. The segment AE divides the rectangle into a right triangle and a trapezoid. The area of the trapezoid is known to be a right angle How many centimeters is the circumference of trapezoid than that of triangle?

From the meaning of the question, point E is on the trisection of CD, then CE = 12 / 3 = 4
Then the perimeter of trapezoid = AE + 8 + 12 + 5
Perimeter of triangle = AE + 4 + 5
The difference is 16

Divide it into a right angle and a trapezoid and draw it into an isosceles triangle

What is LZ asking about? How many painting methods are there?
A vertical line passing through two vertexes is perpendicular to the bottom edge and perpendicular to the waist
There are four kinds

Draw a line in the rectangle and divide it into trapezoid and triangle so that their area ratio is 2:1 Two centimeters in length and four centimeters in width

Let the rectangle be ABCD
If a straight line through a does not coincide with the line where ad is located, continuously take AA '= a'a' '= a'a' '= a''A' ''
Connect a '''d
The parallel line passing through a 'do a' ''d crosses AD and E
Trapezoid AECB and triangle CDE are the results
prove:
∵ ABCD is a rectangle
∴AB=CD,AD=BC
∵AA'=1/3AA''',A'E∥A'''D
∴AE=1/3AD,ED=2/3AD
SAECB = 1/2(AE+BC)*AB = 1/2(1/3AD+BC)*AB = 1/2(1/3AD+AD)*AB = 2/3AD*AB
S△CDE = 1/2*DE*CD = 1/2*2/3AD*AB = 1/3AD*AB
∴SAECB：S△CDE = (2/3AD*AB)：(1/3AD*AB) = 2：1

Draw a line in a triangle so that the triangle is divided into two parts of equal area

Find the middle point of the bottom edge, connect the midpoint of the bottom edge with the corresponding vertex of the bottom edge. At this time, the triangle is divided into two parts with equal area
Because at this time, they wait for the bottom and the height