Let AC = B, BC = a, ab = C, CD = h in RT △ ABC, let AC = B, BC = a, ab = C, CD = H It is known that: in RT △ ABC, ∠ ACB = 90 ° CD ⊥ AB is at point D, let AC = B, BC = a, ab = C, CD = H Proof: a triangle with a + B, h, C + h sides is a right triangle

∵ ABC is RT △, ∵ A & sup2; + B & sup2; = C & sup2;
S△=1/2ab=1/2ch ∴ab=ch
∴(a+b)²+h²=a²+b²+2ab+h²=c²+2ab+h²=c²+2ch+h²=(c+h)²
So the conclusion is true

In the triangle ABC, ∠ ACB = 90 °, AC = BC, point D is on side AB, connect CD, and rotate the line CD 90 ° clockwise around point C to CE position, connect AE

What is this for?

In triangle ABC, if angle c = 90 degrees, SINB = 1 / 5, BC = 2, then AB =?

sinB=AC/AB=1/5
AC=AB/5
From Pythagorean theorem
AB²=AC²+BC²
Then AB & sup2; = AB & sup2 / / 25 + 4
AB²=4÷24/25=25/6
AB=5√6/6

In RT △ ABC, ∠ C = 90 °, BC = 4, AC = 3, after rotating △ ABC around point B, point a falls on point a 'on line BC, and point C falls on point C',
What is the value of AA '?
There is another kind of answer: how to ask for the root 10 of 3?

Sorry, there are two situations:
1) Turn clockwise
∠C = 90°,BC = 4,AC = 3
AB & nbsp; = & nbsp; 5 is obtained from Pythagorean theorem
After ABC rotates around B, a is on the line where BC is,
Then a & # 39; B & nbsp; = & nbsp; ab & nbsp; = & nbsp; 5
∴A'C = A'B - BC = 5-4=1
∵AC = 3
According to Pythagorean theorem, AA & # 39; & nbsp; = & nbsp; √ (9 + 1) & nbsp; = & nbsp; √ 10 & nbsp;
2) Turn counter clockwise
Now a & quot; is on the left side of BC
A"B = AB = 5
∴A"C = 9
∵AC = 3
∴AA" = √(81+9) = 3√10

Let AC = B, BC = a, ab = C, CD = h in RT △ ABC, let AC = B, BC = a, ab = C, CD = H It is known that: in RT △ ABC, ∠ ACB = 90 ° CD ⊥ AB is at point D, let AC = B, BC = a, ab = C, CD = H Proof: a triangle with a + B, h, C + h sides is a right triangle