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It is known that as shown in the figure, ad and BC intersect at point O, ∠ cab = ∠ DBA, AC = BD
It is proved that in △ ABC and △ bad, AC = BD, cab = dbaab, SAS, C = D, in △ AOC and △ BOD, AOC = BOD, C = DAC = BD, AAS
As shown in the figure, ad and BC divide ∠ cab and ∠ DBA equally, and ∠ 1 = 2. Try to explore the size relationship between AC and BD, and explain the reason
To push out the format
Certification:
∵∠CAB=2∠1
∠DBA=2∠2
∠ 1=∠2
∴∠CAB=∠DBA
And ∵ 1 = ∠ 2
AB=AB
∴△CAB≌△DBA
∴AC=BD
As shown in the figure, AC ∥ BD, AE and be are equally divided into cab and DBA, and point E is on CD
It is proved that: as shown in the figure, AF = AC is intercepted on AB and EF is connected. In △ CAE and △ FAE, AC = AF, CAE = ∠ faeae = AE, and ≌ △ CAE ≌ △ FAE (SAS), then ∠ CEA = ∠ FEA, and ∠ CEA + ∠ bed = ∠ FEA + ∠ Feb = 90 °, and ∠ Feb = ∠ DEB, ≌ - be bisection, DBA, and ≌ - DBE = ∠ FBE, in △
In the isosceles trapezoid ABCD, ab ∥ DC, ad = BC = 5, DC = 7, ab = 13, point P starts from point a and moves along the axis at a speed of 2 units per second
In the isosceles trapezoid ABCD, ab ∥ DC, ad = BC = 5, DC = 7, ab = 13, point P starts from point a and moves along ad → DC to terminal C at a speed of 2 units per second, while point Q starts from point B and moves along Ba to terminal a at a speed of 1 unit per second, and the movement time is set as T seconds?
When 0 ＜ t ≤ 2, i.e. n is on ad, two cases are discussed
① When ∠ BMQ = 90 °, i.e. m coincides with P, then BM + pf + CF = BM + Nd + CF = 2T + 1 = 4
The solution is t = 1.5s
② When ∠ BQM = 90 ° in the right triangle NQD, nd = t, ∠ ADB = ∠ DBC = 30 °,
∴NQ= 33t．
∵NP= 3∴QP= 3- 33t
In the right triangle BQM, ∠ DBC = 30 ° BM = t
∴QM= 12t
In the right triangle QPM, ∠ QMP = 60 ° QM = 12t
∴QP= 34t
∴ 3- 33t= 34t．
The solution is t = 127S
When 2 ＜ t ＜ 4, ∠ BQM = 90 °
In right triangle BNP, BN = 4-T, ∠ ABC = 60 °,
∴BP= 4-t2,
∴PM=BM-BP=t- 4-t2= 3t-42
In bpq, DBC = 30 ° BP = 4-t2
∴PQ= 3(4-t)6
In QPM, QMP = 60 ° and PM = 3t-42
∴PQ= 3(3t-4)2
So 3 (4-T) 6 = 3 (3t-4) 2,
The solution is t = 1.6s, which is inconsistent with the value range of T,
Therefore, this situation does not hold
In conclusion, when t = 1.5s or 127S, △ BMQ is a right triangle
Among them, the number has changed, and the thinking is the same
In square ABCD, e. f is on BC and CD respectively, and the angle EAF is 45 degrees. It is proved that the area of triangle AEF = the area of triangle Abe + the area of triangle ADF
Extend FD to g, so that DG = be. Obviously, triangle Abe ≌ triangle ADG, because their two right angles are equal. So, ∠ GAD = ∠ BAE, and ∠ BAE + ∠ DAF = 45, so: angle GAF = angle GAD + angle DAF = angle EAF = 45. Also: Ag = AE, AF = AF, then triangle AEF ≌ triangle AGF
Known: as shown in the figure, in square ABCD, point E is the midpoint of edge CD, point F is on edge BC, and BC = 4CF. Proof: Triangle ade is similar to triangle ECF
Certificate: ∵ square ABCD
∴∠C=∠D=90°,AD=CD=BC
∵ e is the midpoint of CD
∴EC/CD = EC/ AD= 1/2,CD=2DE
∵BC=4CF
∴CF= 1/4 BC =1/4 CD= 1/2 DE
∴EC/AD=CF/DE = 1/2
And AD / EC = de / CF = 2
∴△ADE∽△ECF
If the areas of △ CEF △ Abe △ ADF are 3, 4 and 5 respectively, the area s of △ AEF can be obtained
Let s be the area of the rectangle, which is known as ab · be = 8, CE · CF = 6, ad · DF = 10, s-ab · CE = 8 from ab · be = 8, s-ad · CF = 10 from ad · DF = 10, so (S-8) · (S-10) = ab · ad · CE · CF = 6S, so (S-4) (s-20) = 0, so the area of the rectangle is 20, and the area of the triangle AEF is 20-3-4-5 = 8
ABCD is a rectangle. E and F are the points on BC and CD respectively. The area of Abe is 2, the area of CEF is 3, the area of ADF is 4, and the area of AEF is 3
From the known
AB*BE=2*2=(DF+CF)*BE
DF*AD=4*2=DF*(CE+BE)
Namely
DF*BE+CF*BE=4
DF*CE+DF*BE=8
AEF area = (DF + FC) * (CE + EB) - 2-3-4 = 3
In square ABCD, E.F is on BC and. CD respectively, and the angle EAF is 45. It is proved that the area of triangle AEF = the area of triangle Abe + triangle ADF
Because the quadrilateral ABCD is a square, so AB = ad, rotate the triangle ADF 90 degrees around point a, so that point D and B coincide, from point F to point g. there are: Triangle gab, congruent triangle fad, triangle gab area = triangle fad area, BG = DF, Ag = AF, angle bag = angle DAF, angle ABG = angle ADC = 90 degrees
As shown in the figure, in square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 ° to prove s △ AEF = s △ Abe + s △ ADF
This is to extend CD to m, make DM = be, connect am, make DM = be, connect am, quadriquadrilateral, ABCD is square, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\degree= In △ EAF and △ MAF, AF = AF, EAF = mafae = am ≌ EAF ≌ MAF (SAS), s △ EAF = s △ MAF, s △ MAF = s △ DAF + s △ mad = s △ ADF + s △ Abe, s △ AEF = s △ Abe + s △ ADF
RELATED INFORMATIONS
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