As shown in the figure, in △ ABC, ∠ C = 90 ° D is a point on the edge of BC, de ⊥ AB is e, ∠ ADC = 45 °, if de: AE = 1:5, be = 3, calculate the area of △ abd

As shown in the figure, in △ ABC, ∠ C = 90 ° D is a point on the edge of BC, de ⊥ AB is e, ∠ ADC = 45 °, if de: AE = 1:5, be = 3, calculate the area of △ abd

In △ AED, ∵ de ⊥ AB is equal to E,
And ∵ de: AE = 1; 5,
If de = x, then AE = 5x,
According to Pythagorean theorem, ad2 = AE2 + ED2 = (5x) 2 + x2 = 26x2,
∴AD=
26x．
In △ ADC, ? C = 90 °, ADC = 45 °,
∴∠DAC=45°．
According to Pythagorean theorem, ac2 + DC2 = ad2 = 26x2,
∴AC=DC=
13x．
In RT △ bed, ∵ ed = x, be = 3,
BD2 = ED2 + be2 = x2 + 32 = x2 + 9,
∴BD=
x2+9．
In RT △ bed and RT △ BCA,
∵ B is the common angle,
∠BED=∠BCA=90°，
ν Δ bed ∽ △ BCA, and ab = 3 + 5x
∴ED
AC=BD
BA．
That is X
13x=
x2+9
3+5x．
Solving the equation 3 + 5x about X=
13•
x2+9，
The square of the two sides shows: (3 + 5x) 2 = 13 · (x2 + 9),
The result is: 2x2 + 5x-18 = 0,
That is (x-1) (2x + 9) = 0,
∴x1=2 x2=-9
2．
∵x=ED＞0，
∴x=ED=2，AE=5x=10．
∴AB=AE+BE=10+3=13．
∴S△ABD=1
2ED•AB=1
2×2×13=13．

As shown in the figure: in RT △ ABC, ∠ C = 90 °, a = 22.5 °, DC = BC, de ⊥ AB, verification: AE = be

It is proved that: ∵ in RT △ DBC, ∵ C = 90 °, DC = BC,
∴∠BDC=∠DBC=45°，
∵∠A=22.5°，
∴∠ABD=∠BDC-∠A=45°-22.5°=22.5°，
∴∠A=∠ABD，
∴BD=AD，
And ∵ de ⊥ ab,
∴AE=BE．

As shown in the figure, in △ ABC, ∠ C = 90 ° D is a point on the edge of BC, de ⊥ AB is e, ∠ ADC = 45 °, if de: AE = 1:5, be = 3, calculate the area of △ abd

In △ AED, ∵ de ⊥ AB is equal to E,
And ∵ de: AE = 1; 5,
If de = x, then AE = 5x,
According to Pythagorean theorem, ad2 = AE2 + ED2 = (5x) 2 + x2 = 26x2,
∴AD=
26x．
In △ ADC, ? C = 90 °, ADC = 45 °,
∴∠DAC=45°．
According to Pythagorean theorem, ac2 + DC2 = ad2 = 26x2,
∴AC=DC=
13x．
In RT △ bed, ∵ ed = x, be = 3,
BD2 = ED2 + be2 = x2 + 32 = x2 + 9,
∴BD=
x2+9．
In RT △ bed and RT △ BCA,
∵ B is the common angle,
∠BED=∠BCA=90°，
ν Δ bed ∽ △ BCA, and ab = 3 + 5x
∴ED
AC=BD
BA．
That is X
13x=
x2+9
3+5x．
Solving the equation 3 + 5x about X=
13•
x2+9，
The square of the two sides shows: (3 + 5x) 2 = 13 · (x2 + 9),
The result is: 2x2 + 5x-18 = 0,
That is (x-1) (2x + 9) = 0,
∴x1=2 x2=-9
2．
∵x=ED＞0，
∴x=ED=2，AE=5x=10．
∴AB=AE+BE=10+3=13．
∴S△ABD=1
2ED•AB=1
2×2×13=13．

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ad bisects ∠ BAC, intersects BC and D, passes through point D as de ⊥ AB, De is exactly the bisector of ∠ ADB (2) CD = 1 / 2ad = 1 / 2dB

DAC = DAB = ABC = 30 CD = 1 / 2ad = 1 / 2dB can be explained with angle 30

In the triangle ABC, where ∠ a = 90 °, P is the midpoint of AC, PD ┻ BC, D is the perpendicular foot, BC = 9, DC = 3. Find the value of ab fast

AE ┻ BC is done by a, and E is vertical foot
Then Pd / / AE, and P is the midpoint of AB, so CD = de = be3. According to the Pythagorean theorem, AE ^ 2 = AC ^ 2 - (CD + de) ^ 2
In the right triangle BAE, AB ^ 2 = AE ^ 2 + be ^ 2
AB ^ 2 = AC ^ 2-36 + 9 = AC ^ 2-27 (1)
In the right triangle ABC, AB ^ 2 = BC ^ 2-ac ^ 2 (2)
(1) + 2,
2AB^2=54
AB^2=27

As shown in the figure, in the triangle ABC, the angle a = 90 degrees, P is the midpoint of AC, PD is perpendicular to BC, D is the perpendicular foot, BC is equal to 9, DC is equal to 3, and the length of AB is calculated

PC * cos angle c = CD = 3
BC * cos angle c = 9 * cos angle c = AC = 2pc = 2 * 3 / cos angle c
=>(COS angle c) ^ 2 = 6 / 9
=>(sin angle c) ^ 2 = 1 / 3
=>AB = BC * sin angle c = 9 * (Radix 3 / 3) = 3 * Radix 3

In the triangle ABC, a = 90, P is the midpoint of AC, PD is vertical BC D, vertical foot BC = 9, DC = 3, find the length of ab

∵ C is the common angle,  a = ∠ ADP = 90 °
∴△CDP∽△CAB
∴CP/CB=CD/AC
And ∵ CP = AC / 2
ν (AC / 2) / CB = CD / AC, that is, AC / 18 = 3 / AC
AC=3√6
I haven't learned anything similar
BD=9-3=6,CP=AP
In RT △ CDP, PD? 2 = CP? CD? 2 = CP? 2 - 9
In RT △ PDB, Pb? 2 = PD? 2 + BD? 2 = CP? 2 - 9 + 36 = CP? 2 + 27
In RT △ APB, AB? 2 = Pb? AP? 2 = CP? 2 + 27-ap? 2 = 27
In RT △ ACB, AC 2 = BC 2 - AB 2 = 81-27 = 54
∴AC=√54=3√6

As shown in the figure, we know that in the triangle ABC, the angle a is equal to 90 ° and P is the midpoint of AC, PD is perpendicular to D, BC is equal to 9, and DC is equal to 3. Find the length of ab

If AE is perpendicular to BC and E. P as the midpoint of AC, DC is 3, then CE is 6
AC ^ 2 = CE * CB, then AC ^ 2 = 54
If AB ^ 2 + AC ^ 2 = BC ^ 2, then AB ^ 2 = 27

As shown in the figure, it is known that the triangle ABC is an equilateral triangle, am is the height, and P is any point inside (outside) ABC. Draw a vertical line to the three sides, and find the relationship between am, PD, PE and PF Brother, I want to map, but Baidu does not support ah, but your answer is right, but I want the process, the answer I can guess

If P is a point in the triangle, am = PD + PE + PF
If P is a point outside the triangle, then am = PD + pe-pf (big brother, where is your picture? In short, the sum of the point outside the triangle and the two waist vertical lines minus this point to the bottom vertical line is equal to the height on the bottom edge)

As shown in the figure, there is a point P, PE ⊥ AB, PF ⊥ AC, PD ⊥ BC in the equilateral triangle ABC. The vertical feet are e, F, D respectively, and ah ⊥ BC is in h. The formula of triangle area is used to prove that PE + pf + PD = ah

Proof: connect AP, BP, CP,
∵ PE ⊥ AB, PF ⊥ AC, PD ⊥ BC, ah ⊥ BC at h,
∴S△ABC=1
2BC•AH，S△APB=1
2AB•PE，S△APC=1
2AC•PF，S△BPC=1
2BC•PD
∵S△ABC=S△APB+S△APC+S△BPC
∴1
2BC•AH=1
2AB•PE+1
2AC•PF+1
And ab = BC = AC,
PE + pf + PD = ah