1) Proof: Because DE∨BC
So △ADP∽△ABQ, △APE∽△AQC
So: DP/BQ=AP/AQ.
PE/QC = Asia Pacific /AQ
So DP/BQ=AP/AQ.
2) In the right triangle ABC, from the Pythagorean theorem, BC=√2, and the height of the side of BC is √2/2.
Let the side length FG=DG=x of the square DEFG,
Because DE∨BC,
So △ Ade ∽△ACB
Therefore, the height of delta ade side/the height of delta /δABC side =(√2/2-x)/(√2/2).
The solution is x=√2/3.
By DE∨BC, so △AMN∽△AFG,
Therefore, Mn/fg = height on the edge of Δ ade/height on the edge of Δ ABC =DE/BC.
The solution is MN=√2/9.
In the isosceles right triangle ADE, △ ADM △ aen, DM=EN=DE/3=MN.
So Mn 2 = DM * en
2
1) set EF and AC to O.
Because the crease EF passes through the AD edge at point E and the BC edge at point F,
So EF bisects AC vertically
So AE=EC, AF=FC.
So ∠EAC=∠ECA
Because in the rectangular ABCD, AD∨BC
So ∠ East African Community =∠ACB
So ∠ACB=∠ECA
Because EF⊥AC,
So ∠EOC=∠FOC,
And OC is the public edge.
So △ EOC △ FOC
So EC=FC
So AE=EC=CF=FA.
The quadrilateral AFCE is a diamond.
2) From the above, AE=AF= 10,
Let AB=a, BF=b,
From pythagorean theorem, it is concluded that AB 2+BF 2 = AF 2,
Namely: A 2+B 2 = 10 2
(a+b)^2-2ab= 100，
Because (1/2)ab=24, ab= 196.
So a+b= 100+96= 196,
So a+b= 14.
So the perimeter of △ABF is a+b+AF= 14+ 10=24.
3) cross e as EP⊥AD and AC as p,
Because AEP = AOE = 90,
∠EAO is the angle of * *
So △AEO∽△ ape
So AE/AP=AO/AE
So AE 2 = ao * AP
Because AO=AC/2
So AE 2 = (AC/2) * AP.
Is that 2AE? = AC access point
three
1) O is OE⊥AB, and the vertical foot is e,
In the right triangle BOE, OE=BO/2= 1,
So BE=√3,
From the vertical diameter theorem, AB=2BE=2√3.
2) even ao,
Because AO=DO,
So ∠ Dao = ∠ D = 20.
Similarly ∠BAO=∠B=30.
So ∠BOD=∠BAO+∠DAO=30+20=50.
3) because in △ACD, ∠ CAD >; ∠BAC=30
So it can only be ∠ d = ∠ b = 30.
BO=DO=2
So BO can't be the edge corresponding to DO,
Therefore, when AD/BO=DO/BC is satisfied, △AOD∽△BOC.
So 2√3/2=2/BC,
The solution is BC=(2/3)√3.
At this time AC=(4/3)√3.
four
1) Because the circumferential angles of the AD arc are ∠ABD and ∠ACD.
So ∠ABD=∠ACD
Because the CD bisects the outer corner of ∠ACB.
So ∠ ACD = ∠ DCM (extension line from BC to M)
So ∠ABD=∠DCM
And a, b, c, D*** cycles.
So ∠DCM=∠BAD (the outer angle of a quadrilateral inscribed in a circle is equal to the inner diagonal)
So ∠BAD=∠ABD
So DB=AD
So △ABD is an isosceles triangle.
In progress. . .