In △ ABC, Ba = BC, ∠ ABC = 45 °, ad is the height on the edge of BC, e is a point on ad, ed = CD, connecting EC

In △ ABC, Ba = BC, ∠ ABC = 45 °, ad is the height on the edge of BC, e is a point on ad, ed = CD, connecting EC

Because Ba = BC, ∠ ABC = 45 ° so ∠ BCA = ∠ BAC = 135 ° / 2 = 67.5 ° and ad is vertical to BC, then ∠ ADC = 90 ° and ED = CD shows ∠ ECD = 45 ° so ∠ ace = 67.5 ° - 45 ° = 22.5 ° because ∠ AEC = ∠ EDC + ∠ ECD = 135 ° then ∠ EAC = 180 ° - 22.5 ° - 135 ° = 22.5 ° so EA = EC is proved

It is known that in △ ABC, Ba = BC, ∠ ABC = 45 °, ad is the height on the edge of BC, e is the point on ad, ed = CD, connecting EC
Elder brothers and sisters, please help me think about how to do it. I have urgent need, and I will thank you very much

∠ABC=45°,∠BAD=45°,∠ADC=90°,
Ed = DC, therefore, ∠ Dec = ∠ DCE = 45 °,
So ∠ ECA = 22.5 °, EA = EC

As shown in the figure, in △ ABC, ∠ ACB = 90 °, be bisection ∠ ABC, CF bisection ∠ ACB, CF, be intersection point P, AC = 4cm, BC = 3cm, ab = 5cm, then the area of △ CPB is ()
A. 1cm2B. 1.5cm2C. 2cm2D. 2.5cm2

∵ be bisects ∠ ABC, CF bisects ∠ ACB, and the distance from point P to AB, BC and AC is equal. Let h, ∵ s △ ABC = 12ac · BC = 12 (AB + BC + AC) · h, that is, 12 × 4 × 3 = 12 (5 + 3 + 4) · h. The solution is h = 1, and the area of ∵ △ CPB = 12 × 3 × 1 = 1.5cm2

In the triangle ABC, ab = 10, BC = 9, AC = 17, find the height on the side of BC
You don't need cos or sin. You didn't learn at all

ABC is an obtuse triangle, the height of BC is on its extension line, extend CB, make AE through a, perpendicular to CB extension line, intersect e point, let CE = a, be = B, then in AEB and AEC, according to Pythagorean theorem, AE & sup2; + be & sup2; = AB & sup2;, AE & sup2; + EC & sup2; = AC & sup2; that is a & sup2; + B & sup2; = 10

As shown in the figure, in △ ABC, ab = 10, BC = 9, AC = 17

Extend CB, make ad ⊥ CB extension line and D point, let ad = x, BD = y, AB2 = x2 + Y2 in the right angle △ ADB, ac2 = x2 + (y + BC) 2 in the right angle △ ADC, solve the equation y = 6, x = 8, that is, ad = 8, ∵ ad is the height on the BC side, and the height on the BC side is 8

As shown in Figure ABC, ad is perpendicular to BC and D AB = 13, AC = 8?
Figure. Is the combination of two right triangles into a large triangle

BD^2=AB^2-AD^2
DC^2=AC^2-AD^2
So BD ^ 2-DC ^ 2 = AB ^ 2-ac ^ 2 = 105

In the triangle ABC, a = 2 times the root 3, B = 6, a = 30 degrees, find B

The sine theorem is as follows
a/sinA=b/sinB
2 radical 3 / (1 / 2) = 6 / SINB
SINB = 3 / (2 radical 3) = radical 3 / 2
Since b > A, b > 30 degrees
So, angle B = 60 degrees or 120 degrees

In the triangle ABC, what is the length of AB if ∠ a = 30 degrees, tanb = 1 / 3, BC = √ 10

Make CD ⊥ AB at point D
∵tanB=1/3
∴BD =3CD
∴BC=√10CD
∵BC =√10
∴CD =1
∴BD=3
∵∠A=30°
∴AD=√3
∴AB =3+√3

In triangle ABC, ∠ C = triangle ABC, ∠ C = 90 ° AC = 6. Tanb = 3 / 4,
Let be = x and the area of triangle be y
1) Find the function relation of Y with respect to X and write out the value range of the independent variable
2) If the circle with the diameter of line segment BC is tangent to the circle with the diameter of line segment AE, the length of line segment be is calculated
3) If the triangle with the vertex of BEF is similar to the triangle bed, calculate the area of the triangle bed

(1) ∵ in △ ABC, ∠ C = 90 °, AC = 6, tanb = 3 / 4,
∴BC=8,AB=10,
∴CD=DB=4．
Through point E, eh ⊥ CB is set at h
Then eh = 3x / 5 can be obtained
Y = 1 / 2 × 4 × 3x / 5 = 6x / 5 (0 ＜ x ≤ 16 / 5 or 5 ＜ x ≤ 10)
(2) Take the midpoint o of AE, make og ⊥ BC at g through O, and connect OD
Then og = 3 / 5, OB = 3 / 5 × (10 + x) / 2 = 3 (10 + x) / 10, Gd = cd-cg = 4-2 / 5 (10-x) = 2x / 5,
∴OD= √[9/100(10+x)^2+4x^2/25]．
If two circles are circumscribed, 1 / 2 BC + 1 / 2ae = od can be obtained,
∴（BC+AE）^2=4OD^2,
∴（8+10-x）^2=4[ 9/100（10+x）^2+ 4x^2/25]
The solution is x = 20 / 3
If two circles are inscribed, then | 1 / 2BC - 1 / 2ae | = OD,
∴（BC-AE）^2=4OD^2,
∴（8-10+x）^2=4[ 9/100（10+x）^2+ 4x^2/25]
The solution is x = - 20 / 7 (rounding off), so there is no inscribed circle
Therefore, the length of segment be is 20 / 3
(3) According to the meaning of the title, it can be divided into two cases
① When ∠ bef is an acute angle,
It is known that the triangles with B, e and F as vertices are similar to △ bed, and it is also known that ∠ EBF = ∠ DBE, ∠ bef ＜ ∠ bed, so ∠ bef = ∠ BDE
DM ⊥ Ba over point D and eh ⊥ BC over point e over point M
According to the fact that the remaining angles of equal angles are equal, it can be proved that ∠ MDE = ∠ HDE,
∴EM=EH．
EM = mb-eb = 16 / 5-x,
According to (1): eh = 3x / 5,
∴ 16/5-x=3x/5,
∴x=2．
∴y= 6/5×2=12/5．
② When ∠ bef is an obtuse angle, x-16 / 5 = 3x / 5 can be obtained in the same way
∴x=8．
∴y= 5/5×8=48/5．
So the area of △ bed is 12 / 5 or 48 / 5

In the triangle ABC, ab = AC, BC = 6, tanb = 4 / 3, Sina

Through point a, ad is perpendicular to D. because AB = AC, triangle ABC is isosceles triangle, so ad is the middle perpendicular of isosceles triangle ABC, so BD = DC = 1 / 2BC angle, ADB = 90 degree, because BC = 6, tanb = ad / BD = 4 / 3, so ad = 4, BD = 3AB ^ 2 = ad ^ 2 + BD ^ 2, so AB = 5, so SINB = ad / AB = 4 / 5, according to sine theorem: A /