In the triangle ABC, the angle c = 90 degrees, AC = BC, P is any point of AB, PE is vertical BC, PF is vertical AC, M is the midpoint of AB, connecting me, MF It is proved that me = MF

In the triangle ABC, the angle c = 90 degrees, AC = BC, P is any point of AB, PE is vertical BC, PF is vertical AC, M is the midpoint of AB, connecting me, MF It is proved that me = MF

Establishing Cartesian coordinate system
Let C (0,0) a (0, x) B (x, 0) m (x / 2, X / 2) P (a, x-a)
E(a,0) F(0,x-a)
therefore
MF^2=(x/2)^2 + (x/2-(x-a))^2
ME^2=(x/2-(x-a))^2 + (x/2)^2
So MF = me

The triangle ABC angle a = 90, ab = AC, M is the midpoint of BC, P is any point on BC, PE is perpendicular to AB, PF is perpendicular to AC to F, and me = MF

Firstly, ab = AC, am ⊥ BC are used to calculate am = BM, ∠ B = 45 = ∠ cam
Then PE ⊥ AB, ∠ B = 45. Be = PE is obtained
Then connect AP and use ASA to find △ ape ≌ △ PAF, that is, PE = AF, so AF = be
Finally, △ MBE ≌ △ MAF is obtained by SAS, then me = MF

In △ ABC, AC = BC, M is the midpoint of AB, P is the point where AB is different from M. through P, PE ⊥ AC in E, PF ⊥ BC in F, connect me and MF, and prove me = MF Will use the knowledge about parallelogram and translation of auxiliary line skills! Come on!

MH ⊥ AC, Mg ⊥ BC, Mr ⊥ PF, and PQ ⊥ MH are made by passing M
Then MH = mg, PQ = Mr = FG = EH
So ⊿ MHE ≌ MGF (SAS)
ME=MF

BP CP is bisector of ∠ ABC and ∠ ACB respectively, and PD is parallel to AB, PE is parallel to AC triangle, PDE perimeter is 8cm, find the length of BC

Because BP is the bisector of ABC
So ∠ APB = ∠ CBP,
Because PD ∥ AB,
So ∠ APB = ∠ BPD,
So BD = DP,
Similarly, PE = EC,
Because the circumference of △ PDE = PD + PE + de = 8,
So BD + de + EC = 8,
BC = 8cm

In the triangle ABC, ab = AC, BP is the center line of the triangle ABC, and BP divides the circumference of the triangle into two parts, 20 cm and 18 cm

This problem has to be solved by drawing and calculating by yourself
AB = AC, BP is the middle line of triangle ABC -- "AP = CP = 1 / 2Ab or 1 / 2Ac
BP divides the circumference of triangle into two parts: 20 cm and 18 cm -- > AB + AP = 20cm BC + CP = 18cm (1)
Or AB + AP = 18cm BC + CP = 20cm (2)
From (1) or (2), the length of AB AC can be obtained
（1）AB=40/3 -->BC=18-20/3
(2) AB=12 --->BC=20-6=14

As shown in the figure, triangle ABC is an equilateral triangle, and P is the bisector of angle ABC. A point PE on BD is perpendicular to the vertical of line BP of point E As shown in the figure, △ ABC is an equilateral triangle, P is a point on the bisector BD of ∠ ABC, PE ⊥ AB is at point E, the vertical bisector of line BP intersects BC at point F, and the perpendicular foot is point Q. if BF = 2, then the length of PE is () (proved by Pythagorean theorem)

Solution 1: because BD is the angular bisector of an equilateral triangle. According to the three lines in one, we can know that: BQ = BF * cos30 ° = 2 * √ 3 / 2 = √ 3 so: BP = 2BQ = 2 √ 3 so: PE = BP / 2 = 2 √ 3 / 2 = √ 3 so: PE = √ 3 solution 2 (complex, not recommended): connecting PF, crossing point

As shown in the figure, in RT triangle ABC, CD is the angular bisector of right angle c, e is the midpoint of AB, PE is perpendicular to AB, and the extension of CD at P is proved to be a right triangle As shown in the figure, in RT triangle ABC, CD is the angular bisector of right angle c, e is the midpoint of AB, PE is perpendicular to AB, and CD extension line at P is proved to be isosceles right triangle I've got a wrong number. I think ABP is an isosceles triangle

∵ PE vertical bisection AB, ᙽ PA = Pb
PF ⊥ CB in F, PG ⊥ AC in G
The quadrilateral GPFC is square
∠GPF=90°
△APG≌△BPF
∠APG=∠BPF
Therefore, APB = 90 degrees
So △ ABP is an isosceles right triangle

As shown in the figure, there is a point P, PE ⊥ AB, PF ⊥ AC, PD ⊥ BC in the equilateral triangle ABC. The vertical feet are e, F, D respectively, and ah ⊥ BC is in h. The formula of triangle area is used to prove that PE + pf + PD = ah

Proof: connect AP, BP, CP,
∵ PE ⊥ AB, PF ⊥ AC, PD ⊥ BC, ah ⊥ BC at h,
∴S△ABC=1
2BC•AH，S△APB=1
2AB•PE，S△APC=1
2AC•PF，S△BPC=1
2BC•PD
∵S△ABC=S△APB+S△APC+S△BPC
∴1
2BC•AH=1
2AB•PE+1
2AC•PF+1
And ab = BC = AC,
PE + pf + PD = ah

In the triangle ABC, point P is any point in the triangle, PD is perpendicular to point D, PE is perpendicular to point E, PF is perpendicular to point F, AB is perpendicular to point F (1) Verification: PD + PE + PF is the fixed value; (2) The fixed value is obtained

(1) Proof: connect PA, Pb, PC
The area of △ ABC = △ APB area + △ APC area + △ BPC area
∵∠A＝∠B＝∠C
∴AB=AC=BC
Through a, Ag is perpendicular to BC and G,
So: 1 / 2 Ag · BC = 1 / 2 (PD + PE + PF) · BC
∴PD+PE+PF=AG
AG is the height of the equilateral triangle and is a constant value
So PD + PE + PF is the fixed value
(2) Because △ ABC is an equilateral triangle, ab = a
∴BG=1/2 a
According to the Pythagorean theorem, it can be calculated that Ag = √ 3 / 2

It is known that the point P is inside the equilateral triangle ABC, PD is perpendicular to AB, PE is perpendicular to BC, and PF is perpendicular to CA to F. it is proved that PD + PE + PF is a constant value

If AP, BP and CP are connected, the equilateral triangle ABC consists of three small triangles
If the side length of an equilateral triangle is a and its area is s, then
S=S(ABP)+S(BCP)+S(CAP)
=(1/2)×AB×PD+(1/2)×BC×PE+(1/2)×CA×PF
=(a/2)×PD+(a/2)×PE+(a/2)×PF
=(a/2)×(PD+PE+PF)
So PD + PE + pf = 2S / A
Because s and a have nothing to do with the position of P
So PD + PE + pf = constant value