Junior 3 mathematics, similar to triangles. Hurry, hurry. Question 10

Connect EF to AC

Because cp is perpendicular to de

So CPE= CPD=90 degrees.

So RT triangle CPE vs RT triangle CPD

Because dcp+ ecp=90 degrees.

And because dcp+ PDC=90 degrees.

So cep= dcp

Because CPE= CPD=90 degrees.

So the triangle CPE is similar to the triangle CDP (the two inner angles are equal), so PE:PC=CE:CD

So CPE= DPC

Because of the square ABCD

So DCE=90 degrees = PCE, AD=CD, and AD is parallel to BC because AD=BC and CE=CF

And because AD is perpendicular to CD and CF is perpendicular to CE

So the isosceles rt triangle adc and the isosceles rt triangle cef so the triangle adc is similar to the triangle cef (the two inner angles are equal) so ef:ac = cf:cd

So ef:ac=pe:cp

Because of the isosceles RT triangle ADC and the isosceles RT triangle CEF, DCA= FEC=45 degrees.

Because cep= dcp

So fep= acp

Because ef:ac=pe:cp

So the triangle ACP is similar to the triangle FEP

So apc= epf

So fpa= cpe

Because cp is perpendicular to de

So CPE=90 degrees.

So fpa=90 degrees.

So the PA is perpendicular to the PF

If it's right, I've been playing for a long time

1. From the Pythagorean theorem, ba:ac:bc=5:3:4 because bc=8

Solution: Let cpq cba after x seconds elapse.

Because of the CPQ CBA

So pc:bc=qc:ac

So (8-2x): 8=x:6

The solution is x=2, which is obtained by the Pythagorean theorem, ba:ac:bc=5:3:4 because bc=8

Solution: Let cpq cab after x seconds elapse.

Because of the cpq cab

So pc:ca=qc:bc

So (8-2x): 6=x:8

Solution x = 32 11

AC=6 is obtained from AC:AB=3:5 and BC=8, i.e., AC:

bc=3:4 cpq cba after x seconds, then there is (ac-x):(bc-2x)=3:

4 solution x=0 in the same way cpq cab (ac-x) :(bc-2x) = 4:3

The solution is x=

（1) ∵pq//ac

bpq△bca

The area of bp pc bq qa x (16 x) and bpq abc area [x (16 x) ] and abc area 1 2 16 [ 10 (16 2) 1 2 16 6

The area of 48 bpq is 48 [x (16 x)] and the area of bpq is the area of apq and the area of bq qa is 48 [x (16 x) ]16 x) x].

48x／（16－x）

i.e. y 48x (16x).

2) APQ is similar to ABP.

If apq abp, then apq abp is pq a, apq pac, pac abp

abp acp, (ab ac) pac acp, ap pc

When AP PC, APQ can be similar to ABP.

(0＜x＜16)

2.If APQ can be similar to ABP.

then aqp= apb

And because aqp= b+ bpq= b+ c apb= c+ cap

So it must be c= cap

bp=bc-pc=bc-ac*ac/bc=16-10*10/16=39/4

Answer: Linking BC, MO, extending AB to P makes AP=2AB, linking CP because AM=CM (the result of the first question).

So AOM=90°=ABC

So AMO and ACB are similar.

So am ao=ac ab

Because 2AO=AC

So am ac=ac 2ab=ac ap

So AC2=AM*AP, and AP=2AB

So AC2=2AM*ab

Proof: 2Connect OM, BC

AMC is an isosceles triangle and AO is the midline.

aom=∠com=90°

The circumferential angle of the diameter is a right angle.

abc=90°

then aom= abc=90°

In AOM and ABC.

aom=∠abc=90°

mao=∠bac

aom∽△abc

am/ac=ao/ab

ac*ao=am*ab

again ao=(1 2)ac

ac*(1/2)ac=am*ab

1/2）ac^2=am*ab

ac^2=2am*ab

The second question is ha.

Connect the OM and BC

Since the triangle ACM is an isosceles triangle.

And OM is the middle line of AC according to the three-wire integration.

So OM is perpendicular to AC i.e. angle MOA = 90 degrees.

The angle abc is the circumferential angle corresponding to the straight line, so it is also equal to 90 degrees, which is equal to the angle moa, and because the angle A is equal to the angle A

So the triangle oma is similar to the triangle abc (the two inner angles correspond equally), so there is oa ab=am ac

oa=1/2ac

So 1 2ac 2=am*ab

ac^2=2am*ab

It is the correspondence of similar triangles that becomes a matter of proportion.

Connecting OM, BC, then the RT triangle oma is similar to the RT triangle CBA, (how to prove the right triangle I don't need to say), because it is similar, so ab ac=ao am, because ao=1 2ac, so ab ac=ac 2am, so ac 2=2am*ab.(You may be able to see it clearly when you write it down with a pen).

(1) Because of AD EF BC, the triangle ADB is similar to the triangle EFB, and the triangle ABC is similar to the triangle AFE, so there is EF AD=BF AB, EF BC=AF AB, so EF AD+EF BC

BF AB+AF AB=(BF+AE) AB=AB AB=1(2) From the above, EF AD+EF BC can be known

1, i.e., (ef*bc+ef*ad) ad*bc=1, so ef*(bc+ad) ad*bc=1, i.e., y*(1-2k) k*k=1, y=k*k (1-2k).

Because ab ac=ae ad

So ab ae=ac ad

Since one of them is 90° (somewhat like a congruent HL proof), so ACD Abe so angular cae = angular eab so ae bisects bac

Triangle ABE and Triangle ACD

Because ab ac=ae ad, according to the Pythagorean theorem, cd eb=ad ae is that the three sides of the triangle abe and the triangle acd are proportional.

So triangular ABE and triangular ACD are similar.

cad = bad, i.e.: ae bisects bac

The triangular ADC and triangular AEB are right-angled triangles.

According to ab ac=ae ad, in rt adc and rt aeb, there are two sides that correspond to each other.

Therefore, the RT ADC is similar to the RT AEB

So cae= eab

i.e. AE bisects BAC

In the triangular ADC and triangular AEB.

Because ae ad=ab ac

And because there are two right angles.

So the two triangles are similar.

Similar triangles correspond to equal, i.e. bae = dac so ae bisects bac

Solution: When ap=a, if the triangle apd is similar to the triangle bpc, then the corresponding edge is proportional, and there is:

1. When AD and PB are the corresponding edges, there is: 2 (7-A)=A 3 solution: A=1,6

2. When AD and BC are the corresponding edges, there is: 2 3=A (7-A) solution: A=5 14

∠a=90=∠b

So there are only two cases:

1.Triangle Pad Triangle PBC

So, PA:PB=AD:BC=2:3

So, pa=14 5

2.Triangle pad triangle CBP

So, pa:cb=ad:bp

So, pa*bp=3*2=6

pa(7-pa)=6

To solve the equation, pa=6 or 1

In summary, PA = 14 5 or 6 or 1

If the AP length is X, the BP length is 7 X

APD is similar to BPC

ap/bp=ad/bc

x/7-x=2/3

Solution x=14 5

Solution: Connecting AE;

EF bisects AD;

ae=de；

dae=∠ade；

bad= cad;

bad+∠eac=∠cad+∠eac=∠dae=∠ade；

b+ bad= ade;

bad+∠eac=∠b+∠bad；

eac=∠b；

eac=∠b；

cea=∠aeb；

ace∽△bae；

ae：ce=be：ae；

ae²=ce*be；

and ed=ae;

ed²=ce*be；