(Mathematics) in RT △ ABC, ∠ B = 90 degrees, BC > AB, BP = ba, BP=BA,BD⊥AC,PE⊥AC In the four line segments ad, BD, De, PE, there is a certain quantitative relationship between some line segments. Please write an equation to express the quantitative relationship (the equation contains 2 or 3 line segments), and explain the reasons for the equation

Make PF ⊥ BD and cross BD to F. it is easy to get that the quadrilateral PEDF is a rectangle. Ed = PF, FD = PE
In RT △ BCP and RT △ abd, BP = Ba, so these two right triangles are congruent. So pf = BD = de BF = ad = bd-fd = bd-pe

RT triangle ABC, angle c = 90 degrees, ab = 15, AC: Ba = 3:4, find the area of AC, BC, triangle ABC

AC = 9 BC = 12 area 54

In RT △ ABC as shown in the figure, ∠ B = 90 ° and point P moves from point B to point a at a speed of 1 cm / s along the edge of Ba; at the same time, point Q also moves from point B to point C at a speed of 2 cm / s along the edge of BC? What is the distance of PQ in cm? (the result is expressed by the simplest quadratic root)

If the area of △ PBQ is 35 square centimeter after x seconds, then Pb = x, BQ = 2x. According to the meaning of the title, we get: 12x * 2x = 35, X1 = 35, X2 = - 35 (minus number), so the area of △ PBQ after 35 seconds is 35 square centimeter. PQ = Pb2 + bq2 = x2 + 4x2 = 5x2 = 5 × 35 = 57

As shown in the figure: in RT △ ABC, the angle ABC = 90 ° and BC < AB, take a point P on the extension line of BC, make BP = Ba, respectively pass through points B and P to make the vertical lines BD and PE of AC, and the vertical feet are D, e. verification: ad = PE + BD O (∩) O thank you

The extension line of PF vertical BD intersects point F, because angle PBD = angle a, BP = AB, angle ACB = angle BPF = angle abd, so triangle abd is all equal to triangle BPF, so ad = BF, because DF = PE, ad = PE + BD

As shown in the figure, in RT △ ABC, ∠ BAC = 90 °, ab = 3, AC = 4, P is the point on BC, PE ⊥ AB is in E, PD ⊥ AC is in D. if BP = x, PD + PE is equal to () A. 4-x Five B. 12x 5−12x2 Twenty-five C. 7 Two D. x 5+3

∵ RT △ ABC, ∵ BAC = 90 °, ab = 3, AC = 4,
From Pythagorean theorem BC is obtained=
AB2+AC2=5，
∵AB⊥AC，PE⊥AB，PD⊥AC，
∴PE∥AC，PD∥AB，
∴△CDP∽△CAB，△BPE∽△BCA
∴PD
BA=PC
BC，PE
AC=BP
BC，
∴PD=3(5−x)
5，PE=4x
5，
∴PD+PE=x
5+3，
Therefore, D

In RT △ ABC, ab ⊥ AC, ab = 3, AC = 4, P is a point on the edge of BC. If P is used as PE ⊥ AB in E, PD ⊥ AB in D, let BP = x, then the value of PD + PE is________

Pythagorean theorem, BC = 5
BEP is similar to BAC, PE / Ca = BP / BC, so PE / 4 = x / 5, PE = 4x / 5
PDC is similar to BAC, Pd / BA = PC / BC, so Pd / 3 = (5-x) / 5 = 1-x / 5, PD = 3-3x / 5
So PD + PE = 3 + X / 5

In △ ABC, BP: PC = 3:4, PE ‖ AB, PD ‖ AC, then s △ ABC: s quadrilateral ADPE is equal to?

∵PE∥AB,PD∥AC
∴△ABC∽△DBP∽△EPC,
And ∵ BC: BP: PC = 7:3:4
∴S△ABC：S△DBP：S△EPC=49:9:16
﹤ s △ ABC: s quadrilateral ADPE = 49: (49-9-16) = 49 / 24

In the triangle ABC, ab = AC = 5, SINB = 3 / 5, P on BC, PD perpendicular to AC, PD crossing AC to D, PE perpendicular PD, PE intersecting AB to e, let BP = x, triangle ped surface If the product is y, find the analytic formula of the function of y about X (2) when BP is the value, the area of triangle PED is the largest. (3) whether there is such a point P that the triangle with PDE as the vertex is similar to PDC. If there is, find BP

Because SINB = 3 / 5
(sinB)^2+（cosB)^2=1
So CoSb = 4 / 5
In triangle ABC, it is obtained from cosine theorem that:
AC^2=AB^2+BC^2-2*AB*BC*cosB
Because AB = AC = 5
So BC = 8
PC=BC-BP=8-x
Angle B = angle c
So SINB = sinc
Because PE is perpendicular to E
So the angle DPE is 90 degrees
Because PD is perpendicular to d
So the angle ADP = angle PDC = 90 degrees
So sinc = Pd / PC
So PD = (24-3x) / 5
Angle DPE + angle APD = 180 degrees
So PE is parallel to AC
So PE is parallel to AC
So PE / AC = BP / BC
So PE = 5x / 8
Because the area of the triangle ped = 1 / 2 * PE * PD = y
(1) So y = - (3 / 16) * (x ^ 2-8x) = - (3 / 16) * (x-4) ^ 2 + 3
(2) When x = 4, the area of triangle PED is the largest
(3) There is a point P such as a triangle, which makes the triangle PDE with P as the vertex similar to the triangle PDC
So PE / PD = Pd / DC or PE / PD = DC / PD
Because DC / PC = COSC = 4 / 5
DC=4/5(8-X)
BP = x = 144 / 23 or 256 / 57

As shown in the figure, in △ ABC, BC = 5cm, BP and CP are angular bisectors of ∠ ABC and ∠ ACB respectively, and PD ‖ AB, PE ‖ AC, then the circumference of △ PDE is () A. 10 B. 15 C. 20 D. 5

∵ BP and CP are the angular bisectors of ∵ ABC and ∠ ACB respectively,
∴∠ABP=∠PBD，∠ACP=∠PCE，
∵PD∥AB，PE∥AC，
∴∠ABP=∠BPD，∠ACP=∠CPE，
∴∠PBD=∠BPD，∠PCE=∠CPE，
∴BD=PD，CE=PE，
The circumference of △ PDE = PD + de + PE = BD + de + EC = BC = 5cm
The circumference of △ PDE is 5cm
Therefore, D

In the triangle ABC, the angle ACB = 90 degrees, AC = BC, M is the midpoint of AB, P is a point on AB, PE is perpendicular to AC, PF is perpendicular to BC, verification: me = MF, me vertical MF

Proof: it is easy to know that ACB, AEP, AMC and BMC are isosceles right triangles, and epcf are rectangles