An isosceles right triangle ABC, angle ACB is 90 degrees. AC equals BC equals 5, there is a point P in the triangle, AP is root 5, BP = BC, find PC!

An isosceles right triangle ABC, angle ACB is 90 degrees. AC equals BC equals 5, there is a point P in the triangle, AP is root 5, BP = BC, find PC!

Hint, the detailed process of their own supplement
PD over P is perpendicular to ab over D, CF over C is perpendicular to ab over F
AP ^ - ad ^ = BP ^ - (ab-bd) ^ (^ = Square)
Solve ad, get DP
DF=AF-AD
PC ^ = DF ^ + (cf-dp) ^

Am is the middle line of △ ABC. Verification: am < 12 (AB + AC)

It is proved that: extending am to point D, making MD = am, connecting BD, it is easy to prove that △ AMC and △ BMD are congruent,  BD = AC, in △ abd, ad ＜ AB + BD, ﹤ 2am ＜ AB + BD, ﹤ 2am ＜ AB + AC, ﹤ am ＜ 12 (AB + AC)

Take a point P in the isosceles right triangle ABC (angle c = 90 degrees), and AP = AC = a, BP = CP = B, and prove that a ^ 2 + B ^ 2 / A ^ 2-B ^ 2 is the fixed value

Because AP = AC = a, P takes a as the center of the circle and the length of AC as the original point on the radius (that is, the radius is a). Because BP = CP = B, P has two points on the vertical line of BC (discussed separately, and a is the fixed value at this time). There is something wrong with the formula A ^ 2 + B ^ 2 / A ^ 2-B ^ 2 = a ^ 2 + (b ^ 2-A ^ 2 × B ^ 2) / A ^ 2 = a ^ 2 + B ^ 2

Triangle ABC, ab = 8, AC = 6, am is the middle line of BC, then what is the value range of am?

Extend am to e so that am = me, AE = 2am
Because BM = MC, AMC = BME
Triangle AMC congruent triangle BME
BE=AC
BE-AB

In the triangle ABC, angle a = 80 degrees. If the distance between a point m in the triangle and three vertices is equal, then angle BMC =?

The problem is simple: point m is the outer center of triangle ABC, the angle BMC is the center angle of inferior arc BC, and the circular angle of inferior arc BC is angle a, that is, BMC = 2 * angle a = 160 degrees

If △ ABC three sides a, B, C satisfy a & # 178; + B & # 178; + C & # 178; = 12a + 16b + 20c-200, what is the area of △ ABC?

Certification:
∵a2+b2+c2=12a+16b+20c-200
(a-6) square + (B-8) square + (C-10) square = 0
∴a=6 b=8 c=10
∵ 6 square + 8 square = 10 square
This triangle is a right triangle
∴S△ABC=（6×8）÷2=24

If ABC on three sides of ABC satisfies the absolute value of a + 2b-18 + (B-18) ^ 2 + c-30 = 0, the shape of triangle ABC is judged
If ABC satisfies the absolute value of a + 2b-18 + (B-18) ^ 2 + c-30 = 0
Judge the shape of triangle ABC
The key is the absolute value of a + 2b-18, a is negative!
a=-18!
But the side of a triangle can't be negative! What should I do!
Give the process or help solve the problem directly. Thank you

Because the absolute value of a + 2b-18 + (B-18) ^ 2 + c-30 = 0
therefore
a+2b-18=0,(b-18)^2=0,c-30=0
therefore
A + 2B = 18
b=18
c=30
thus
a=

If the three sides a, B and C of triangle ABC satisfy the following formula (a + 2b-60) with square + | B-18 | + | c-30 | = 0, what is the shape of the triangle?

The absolute value of the sum of squares must be greater than or equal to 0, so to make the relationship hold, only a + 2b-60 = 0, B-18 = 0 and c-30 = 0, so a = 24, B = 18 and C = 30 conform to Pythagorean theorem, so they are right triangles

If the trilateral lengths a, B and C of △ ABC satisfy the relation (a + 2b-60) ^ 2 + | B-18 + | c-30 | = 0, then △ ABC is a right triangle and the maximum angle is ∠ C. why?

∵ (a + 2b-60) ≥ 0, | B-18 | ≥ 0, | c-30 | ≥ 0 ∵ a + 2b-60 = 0, B-18 = 0, c-30 = 0 the solution is: a = 24, B = 18, C = 30 ∵ a + B = C ∵ ABC is a right triangle, ∠ C = 90 ° in fact, just need to note that the absolute value of the sum of squares is non negative, but their sum is 0, which means that every term is 0, and then

Excuse me: if the square of parabola y = x - 3x + 2 intersects with X axis at points a and B, and intersects with y axis at point C, then the area of triangle ABC is

According to the meaning of the title, when the parabola and y-axis intersect at the point, x = 0, y = 2, draw a graph, you will know that in the triangle cob, its area is equal to 2, and the area of - COA is the area of ABC