As shown in the figure, in △ ABC, ∠ C = 90 °, AC = 12cm, ab = 25cm, point D on BC, de ⊥ AB, perpendicular foot e, and de = DC, then be=______ .

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = 12cm, ab = 25cm, point D on BC, de ⊥ AB, perpendicular foot e, and de = DC, then be=______ .

In RT △ ACD and RT △ AED,
AD＝AD
DE＝DC ，
∴Rt△ACD≌Rt△AED（HL），
∴AE=AC=12cm，
∴BE=AB-AE=25-12=13cm．
So the answer is: 13cm

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = 12cm, ab = 25cm, point D on BC, de ⊥ AB, perpendicular foot e, and de = DC, then be=______ .

In RT △ ACD and RT △ AED,
AD＝AD
DE＝DC ，
∴Rt△ACD≌Rt△AED（HL），
∴AE=AC=12cm，
∴BE=AB-AE=25-12=13cm．
So the answer is: 13cm

As shown in the figure, in the triangle ABC, BD is the angular bisector, De is perpendicular to point E, and ab = 18cm, BC = 12cm, and the area of triangle ABC is 30 cubic centimeter, then De= If the area of the triangle ABC is 36 cubic centimeters, then de=

DF is perpendicular to F, and BD is an angular bisector, so DF = De
So the area of the triangle ABC is 1 / 2 × (12 + 18) × de = 30
So de = 2

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = 12cm, ab = 25cm, point D on BC, de ⊥ AB, perpendicular foot e, and de = DC, then be=______ .

In RT △ ACD and RT △ AED,
AD＝AD
DE＝DC ，
∴Rt△ACD≌Rt△AED（HL），
∴AE=AC=12cm，
∴BE=AB-AE=25-12=13cm．
So the answer is: 13cm

As shown in the figure, in the triangle ABC, the angle c = 90 degrees, AC = 12cm, ab = 25cm, point D is on BC, De is perpendicular to AB, perpendicular foot is e, and de = DC, triangle bed The ratio of AED area to triangle AED area is 0

13:12

In the triangle ABC, ab = AC, point D is on BC, and ad = BD, DC = AC, calculate the degree of angle B

Answer: 36
If the angle B is x, then the angle bad and angle c are both X (isosceles triangle ADB and isosceles triangle BAC), then the angle DAC is (180-x) / 2. From the sum of inner angles of triangle BAC is 180, x + X + (x + (180-x) / 2) = 180
X = 36

As shown in the figure, in △ ABC, ad is the height on BC, tanb = cos ∠ DAC (1) Confirmation: AC = BD; (2) If sin ∠ C = 12 13, BC = 12, find the length of AD

(1) It is proved that: ∵ ad is the height of BC,
∴AD⊥BC，
∴∠ADB=90°，∠ADC=90°，
In RT △ abd and RT △ ADC,
∵tanB=AD
BD，cos∠DAC=AD
AC，
And ∵ tanb = cos ∠ DAC,
∴AD
BD=AD
AC，
∴AC=BD．
(2) In RT △ ADC, sinc = 12
13，
So we can set ad = 12K, AC = 13K,
∴CD=
AC2−AD2=5k，
∵ BC = BD + CD, and AC = BD,
∴BC=13k+5k=18k
From the known BC = 12,
∴18k=12，
∴k=2
3，
∴AD=12k=12×2
3=8．

As shown in the figure, in △ ABC, ad is the height on BC, tanb = cos ∠ DAC (1) Confirmation: AC = BD; (2) If sin ∠ C = 12 13, BC = 12, find the length of AD

(1) It is proved that: ∵ ad is the height of BC,
∴AD⊥BC，
∴∠ADB=90°，∠ADC=90°，
In RT △ abd and RT △ ADC,
∵tanB=AD
BD，cos∠DAC=AD
AC，
And ∵ tanb = cos ∠ DAC,
∴AD
BD=AD
AC，
∴AC=BD．
(2) In RT △ ADC, sinc = 12
13，
So we can set ad = 12K, AC = 13K,
∴CD=
AC2−AD2=5k，
∵ BC = BD + CD, and AC = BD,
∴BC=13k+5k=18k
From the known BC = 12,
∴18k=12，
∴k=2
3，
∴AD=12k=12×2
3=8．

In the triangle ABC, ad is the height on the edge of BC, tanb = cos ∠ DAC, (1) prove AC = BD

It is proved that because tanb = cos ∠ DAC
And tanb = SINB / CoSb = (AD / AB) / (BD / AB) = ad / BD
cos∠DAC=AD/AC
So ad / BD = ad / AC
AC = BD

As shown in the figure, in △ ABC, ad is the height on BC edge, tanb = cos ∠ DAC (1) verification: AC = BD (2) if sinc = 12 / 13, BC = 12, find the length of AD

（1）tanB=AD/BD,cos∠DAC=AD/AC
∵tanB=cos∠DAC
∴AD/BD=AD/AC
So AC = BD
（2）AD=8
cos∠DAC=sinC=12/13=tanB
Therefore, AD / BD = ad / AC = 12 / 13
Let ad = 12x, then BD = 13X, AC = 13X
In a right triangle ADC, CD = 5x is easily obtained
∴13x+5X=12
Ad = 12x = 8