It is known that in the square ABCD, P is the point inside the shape, APB 135, AB 3, BP 1, find PC

Mr. Wu is going to ask himself again! I'm not young, and I'm still doing some unsavory things.

Solution: Rotate ABP to BCM and connect PM

Apparently BP BM 1, CM PA 3, ABP CBM, BMC APB 135°

So pbm abc 90°

So PBM is an isosceles right triangle.

So PM 2*PB2, PBM 45°, so PMC 135°, 45°, 90°

So the triangle PMC is a right triangle.

According to the Pythagorean theorem: pc 2 pm 2 cm 2 2 + 3 = 5 so pc 5

Jiangsu Wu Yunchao answer for reference!

It is known that in the square ABCD, P is one point in the shape, APB=135°, AB= 3, BP=1, find PC

Analysis: In ABP, APB=135°, AB= 3 is determined by the cosine theorem AB 2=BP 2+AP 2-2BP*AP*COS135==>AP 2+ 2AP-2=0==>AP=( 10- 2) 2

Rotate ABP, with B as the center, 90 degrees clockwise to obtain BCP'

abp≌⊿bcp’

then bp'=90°,bp=bp'==>pp'= 2pp'b=45°==> pp'c=135-45=90°pc 2=2*bp 2+ap 2==>pc= (5- 5).

Summary. Hello <>

The length of the PQ is 5-1. The solution is as follows: from the meaning of the question, it can be seen that the triangle abp is an isosceles right triangle, so pab=45 degrees, pba=45 degrees.

And because bp=ab= 5, the triangle pab is an isosceles right triangle with a side length of 5, so pa=pb= (5 2+ 5 2)= 2*5)= 10. Now consider the triangle PBQ, because PB=BQ, and since PBA= PBC, the triangle PBC is congruent with the triangle PBQ, so QC=PC. And because bp=ab= 5, aq=pa-ab= 10- 5.

Thus, the length of pq can be obtained by the Pythagorean theorem as (pc 2+qc 2)= pa 2-aq 2-ab 2+qc 2)= 10-5-5+pc 2)= 5-1. This question involves geometric knowledge and the Pythagorean theorem, and requires familiarity with the properties of the isosceles right triangle of triangles, the properties of triangle congruence, and the application of the Pythagorean theorem. In addition, it is necessary to understand the concept and nature of the bisector.

When solving problems, you can deepen your understanding of geometry by trying to use other geometric theorems such as the sine theorem or cosine theorem to verify them. <>

The bisector bq crosses the extension of AP at the point Q, and if AP 2, find the length of PQ.

Hello teacher, can you see the meaning of the question?

Hello <>

The length of the PQ is 5-1. The idea of solving the problem is as follows: from the meaning of the question, it can be seen that the triangle segment abp is an isosceles right triangle, so pab=45 degrees, pba=45 degrees.

And because bp=ab= 5, the triangle pab is an isosceles right triangle with a side length of 5, so pa=pb= (5 2+ 5 2)= 2*5)= 10. Now consider the triangle PBQ, because the macro PB=BQ, and since PBA= PBC, the triangle PBC is congruent with the triangular focal PBQ, so QC=PC. And because bp=ab= 5, aq=pa-ab= 10- 5.

Thus, the length of pq can be obtained by the Pythagorean theorem as (pc 2+qc 2)= pa 2-aq 2-ab 2+qc 2)= 10-5-5+pc 2)= 5-1. This question involves geometric knowledge and the Pythagorean theorem, and requires familiarity with the properties of the isosceles right triangle of triangles, the properties of triangle congruence, and the application of the Pythagorean theorem. In addition, it is necessary to understand the concept and nature of the bisector.

When solving problems, you can deepen your understanding of geometry by trying to use other geometric theorems such as the sine theorem or cosine theorem to verify them. <>

The answer is that the length of the LPQ is 5-1

Why is ABP an isosceles right triangle?

It is not a right triangle.

This problem should have a graph, and you didn't provide a graph for it.

Wait for me to take a picture and send it to Yuu. Good.

Let Tongzi dig ab=a

b(0,0),c(a,0),d(a,a),a(0,a) with a,b,c as the center of the circle, and the circle with radius of 1,2,3 intersects the p-point equation of the local core x 2 + y 2 = 4

x 2 + y-a) 2 = 1 --2)x-a) 2 + y 2 = 9 --3)2)+(3) gives (x-a) 2 +(y-a) 2 +x 2+y 2 = 10

or (x-a) 2 + (y-a) 2 = 6, i.e. p is on a circle with d(a,a) as the center and the root number (6) as the radius, so pd = root number (6).

Solve the above equations.

x=(a^2-5)/(2a)

y = a^2+3)/(2a)

a = sqrt(5+2sqrt(2))

Apply the cosine theorem in APB.

a^2 = 5-4cosapb

cosapb = sqrt(2)/2

apb = 135 degrees.

Summary. Kiss <>

I'm glad to answer for you, as shown in the figure, in the square ABCD, AB=root number 5, P is a point in the square ABCD, and BP=AB, connect AP, PBC's bisector BQ crosses AP's extension line at the point Q, if AP 2, find the length of PQ is 1

As shown in the figure, in the square ABCD, AB=root number 5, P is a point in the square slag front crack ABCD, and BP=AB, connecting AP, PBC is flat as the extension of the closed line BQ crosses AP at the point Q, if AP 2, regret contains the length of PQ.

Okay, teacher.

The answer to this question is pq 1, I don't know how to do it, so I trouble the teacher to write down the process of solving the problem.

Kiss [watch the barrier and touch the reputation] <>

I'm glad to answer for you, as shown in the figure, in the square ABCD, AB=root number 5, P is a point in the square ABCD, and BP=AB, connecting AP, PBC's bisector BQ crosses AP's extension section of the long line at the point Q, if AP 2, find the length of PQ is 1

I don't understand. How to get pq l, I want to see the process.

Thank you, teacher! You can click on my avatar to follow Oh, and if you have any questions, feel free to find the teacher

Rotate APB 90° clockwise around point B and connect PE, rotate APB 90° clockwise around point B to obtain BEC, BEC BPA, APB= BEC, source rot and hail return BEP is an isosceles right-angled calendar triangle, BEP=45°, PB=2, PE=22, PC=3,CE=PA=1, PC2=PE2+CE2, PE=.

<> rotate the PAB 90° clockwise around point B to obtain P BC, then Pab P BC, let Pa = X, Pb = 2X, PC = 3X, with PP, isosceles right angle Pbp , P 2 = (2X) 2 + (2X) 2 = 8X 2, PP B = 45°

and pc 2 = pp 2 + p c 2 , get pp c = 90° so slow cherry juice apb= cp disturbance b = 45 ° + 90 ° = 135 ° so the answer is: 135 °

Proof that: (1) the quadrilateral ABCD is a square, bc=ab, (1 point).

cbp=∠abe，bp=be，∴△cbp≌△abe．

Proof of: (2) CBP= ABEs, PBE= ABE+ ABP= CBP+ ABP=90°, PBbe

1) and (2) the two sub-questions can be proved together

Proof: CBP= ABE, PBE= ABE+ ABP (1 point).

cbp+∠abp

90° (2 minutes).

PB be (3 points).

With B as the center of rotation, rotate CBP 90° (4 minutes) in a clockwise direction

BC=AB, CBA= PBE=90°, BE=BP (5 points).

CBP coincides with ABE, CBP ABE (6 points).

Solution: (3) Connect PE, BE=BP PBE=90°, BPE=45°, (7 points).

If ap is k, bp=be=2k, pe2=8k2, (8 points).

PE=2 2K, BPA=135°, BPE=45°, APE=90°, (9 points).

AE = 3K, in right-angle APE: cos PAE = APAE = 13 (10 points).