Original proposition: √ a ＞ B, then a ＞ B (write its inverse proposition, no proposition, inverse no proposition)

Original proposition: √ a ＞ B, then a ＞ B (write its inverse proposition, no proposition, inverse no proposition)

There is an implicit condition, that is, a > = 0, b > = 0,
therefore
If a > b > = 0, then √ a > √ B
No proposition, if √ a

The inverse proposition of the proposition "let a, B, C ∈ R, if ac2 > BC2, then a > b" is______ Proposition (fill in: true or false)

The converse proposition of the proposition "let a, B, C ∈ R, if ac2 ＞ BC2, then a ＞ B" is: let a, B, C ∈ R, if a ＞ B, then ac2 ＞ BC2. When C = 0, the converse proposition does not hold, and the converse proposition of the proposition "let a, B, C ∈ R, if ac2 ＞ BC2, then a ＞ B" is false

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, point D is a point on the hypotenuse AB, make ∠ CDE = ∠ a, cross point C as CE ⊥ CD, cross De to e, connect be. (1) verification: CECB = CDCA; (2) verification: ab ⊥ be

It is proved that: (1) ce ⊥ CD, ∧ DCE = ∧ ACB = 90 ° and ∧ CDE = ∧ a ∧ DCE ∧ ACB, ∧ CECB = CDCA; (2) CECB = CDCA, ∧ CECD = CBCA, ∧ DCE = ∧ ACB = 90 °, ∧ BCE = ∧ ACD, ∧ BCE ∧ ACD, ∧ CBE = ∧ a, ∧ a + ∧ ABC = 90 °, CBE + ∧ ABC = 90 °, Abe = 90 °, ab ⊥ be

As shown in the figure, in RT △ ABC, ∠ ACB is 90 ° and point D is a point on the hypotenuse AB, making ∠ CDE = ∠ a, passing through point C making CE ⊥ CD, crossing De to e, connecting be
Verification: CE than CB = CD than ca
There is no Tula

）∵CE⊥CD,
∴∠DCE=∠ACB=90°
And ∵ ∠ CDE = ∠ a
∴△DCE∽△ACB,
∴ ；

As shown in the figure, rotate the right angle triangle ABC (where ∠ ABC = 60 °) clockwise around point B to the position of a1bc1, so that points a, B and C1 are on the same straight line. If the length of AB is 10, then the distance from point a to point A1 is equal to the distance______ (results π reserved)

∵ - a = 30 °, ∵ C = 90 °, ∵ a 'BC' is obtained by △ ABC rotation, ∵ - ABC = ∵ a 'BC' = 60 °, ∵ ABA '= 180 ° - ∵ a' BC '= 180 ° - 60 ° = 120 °, ∵ AB length is 10, ∵ the distance from point a to point A1 = 120 π· 10180 = 203 π

As shown in the figure, ab = AC, ∠ BAC = 120 ° in △ ABC, ad ⊥ AC intersects BC at point D, and BC = 3aD

It is proved that in △ ABC, ∵ AB = AC, ∵ BAC = 120 °, and ∵ ad ⊥ AC, ∵ DAC = 90 °, and ∵ C = 30 °, CD = 2ad, ∵ bad = ∵ B = 30 °, ad = dB, ∵ BC = CD + BD = AD + DC = AD + 2ad = 3aD

As shown in the figure, it is known that △ ABC is an isosceles triangle, ∠ ACB = 90 ° and through the midpoint D of BC, make de ⊥ AB to e, connect CE and find the value of sin ⊥ ace

Root 2 divided by 2
Because △ ABC is an isosceles triangle, ∠ ACB = 90 degree
Therefore, a = b = 45 degree
Because the midpoint D of BC is de ⊥ AB in E
So △ BDE ≌ △ CDE (SAS)
Therefore, ECD = b = 45 degree
So, ACE = 45 degree
So sin ∠ ace = root 2 divided by 2 (2 / 2 root 2)

As shown in the figure, Bo is the middle line on the hypotenuse of RT △ ABC, extend B0 to point D, so that Bo = do, connect AD and CD, verify: △ AOD congruent △ cob

∵ Bo is the middle line of Δ ABC, ∵ Ao = Co,
∵BO=DO,∠AOD=∠BOC,
∴ΔAOD≌ΔCOB(SAS).