From the topic, BN=NQ, AM=MP, N, M is the midpoint between AC and BC, Mn//QC, Mn//CP.

So QC//CP, so Q, C and P are on the same straight line.

2. Solve the coordinates of C, that is, solve the symmetry point of C about O..

C is (1, 3/2)

Find the coordinate of d as (2,3/2) and let BD be y = kx+b.

Bring in the coordinates of b and d, b=0, k=3/4, and the straight line is y = (3/4) x.

3. Make AG perpendicular to AF, and extend BC to make AG cross with G..

Because ∠GAB+∠BAE+∠EAF=90 degrees, ∠BAE+∠EAF+∠FAD=90 degrees.

So ∠GAB=∠FAD, and because AB=AD, ∠ABG=∠ADF.

So △ABG is equal to △ADF.

So BG=DF, ∠AGB=∠AFD.

Because AB//CD has ∠BAF=∠ADF, ∠AGB=∠BAF.

Because AF is the bisector of ∠EAD.

So ∠GAE =∠ BAFF

So ∠GAE =∠ age, that is, AE=GE=BG+BE=DF+BE.

(1) △ CBD is an equilateral triangle.

The proof is as follows: α+∠HCB=∠HCB+∠BCD=90 degrees.

∠BCD=60 degrees, BC=CD, so) △CBD is an equilateral triangle.

(2) Let the H coordinate be (a, 4)AH=HC.

A squared =(6-a) squared +4 squared.

Get a=3/ 13.

That is, h (3/13,4)

With y=kx+b, the coordinates of h and c can be brought in, and the analytical formula of HC can be obtained.

5. (2) Delta ABC and Delta △CDE are equilateral triangles.

After rotation, a, c and d are on the same straight line.

From the perspective of rotation, BC=AC, CE=CD, and ∠BCA=∠ECD=60 degrees.

Therefore, △ABC and △CDE are equilateral triangles.

(3)△ACN and △BCM are congruent, which is known by rotation.

MN//BD

Because △ACN and △BCM are congruent, there is CM=CN.

Because ∠ACE=60 degrees

So △MCN is an equilateral triangle.

From the perspective of relationship, MN//BD

These are all typed by myself ~ for the landlord's reference ~ ~ The key is to make auxiliary lines and have a conceptual understanding ~ Come on ~