Junior high school math to find the second sub problem, the second sub question of this junior high

BAC = 90°, dm bc (known).

BAC = BMD (Equal Substitution).

b is the common corner (known).

abc∽△bmd（aa）

d= c (property of a similar triangle).

am=mc (the middle line of a right triangle is equal to half of the hypotenuse) mac= c (property of an isosceles triangle).

mac = d (equivalent substitution).

and AMD is a common corner (known).

ame∽△amd(aa)

ae ad = me am (property of a similar triangle) ae ad = me am (property of the equation) ae ad = me md (verified) ae ad = me md

∠bac＝90°，bm＝mc

am＝bm∠b=∠bam

mae = d( b + d = 90 = bam + mae) ame = ame (common angle).

aem∽δdam

ae/ad＝ma/dm

ae ad) 2 (ma dm) 2=ma square md square.

Obtained from the first question.

ma square = md·me

ae ad) 2 (ma dm) 2=ma square md square=md·me md 2=me md

There should be a letter wrong in the question, and the known condition should be: apm= bpn points a, p, b'Whether or not they are on the same straight line.

bp is symmetrical with b'p with respect to mn, so b'pn= bpn

Because, apm= bpn

So, apm= b'pn

It shows that the angles between ap, pb' and mn are equal, and a, b' are on both sides of mn, so the points a, p, b'Whether or not they are on the same straight line.

If you have any questions, please ask.

1, omitted 2, yes. Proof:

Because: bp and b'p on mn symmetry.

So: the angle bpn is equal to the angle b'pn

Because: the angle BPM is equal to the angle APM (known), and the angle BPN is equal to the angle B'pn (proven) So: the angle apm is equal to the angle b'PN (Equivalent Substitution).

Because: angle b'PM and angular B'PN complements (MPN is three-point collinear), and the angle APM is equal to the angle B'PN (verified).

So: angular apm and angular b'PM complementarity (equivalent substitution) so: APB'The three-point collinear certificate is completed.

ABCD is a parallelogram.

So ab=cd=4

ad=bc=6

And because CE=1 2BC

So ce=3

Since abe = 60 degrees and is a parallelogram, dce is also equal to 60 degrees.

Overpower d to do dk be

So cd=2ck=2 [hypotenuse is twice the corresponding edge of 30 degrees] Pythagorean theorem gives dk as 2 times the root number three.

Because ek = ce-ck

So ek=1

Continuing with the Pythagorean theorem, we get that de is the root number thirteen.

I have already given you a caution, and now I will continue to do it.

I forgot about the range of x, and the tan function is a trigonometric function used in high school.

Let me tell you about the general idea of Changpin.

If we take the point D on the PA such that AD=PB, then the SAS theorem proves that ACD is equal to BCP

Therefore, it is necessary to get cd=cp, pb=ad

Whereas apc= abc=60

Therefore, Hengxun and PCD are equilateral triangles, then there is cd=pd, so pa=pd+ad=pc+pb