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In △ ABC, D is a point on the edge of BC, BD = 1 2dc, ∠ ADB = 120 ° ad = 2, if the area of △ ADC is 3, then AB = () A. 1 B. Five C. Seven D. 2 Two
∵∠ADC=π-∠ADB=π
3，
∴S△ADC=1
2•AD•DC•sin∠ADC=1
2•2•DC•
Three
2=
3，
∴DC=2，
∴BD=1
2DC=1，
∴AB=
BD2+AD2−2BD•AD•cos∠ADB=
1+4+2×1×2×1
2=
7．
Therefore, C
1. In the triangle ABC, D is a point on side BC, BD = 1 / 2dc, angle ADB = 120 degrees, ad = 2. If the area of triangle ADC is 3 - √￣ 3, then angle ABC =? 1?
ADB = 120, ADC = 60, ad = 2, s △ ADC = 1 / 2 · ad · DC · sinsinsinadc = 3-3, DC = 3 - √ 3, DC = 3 - √ 3, DC = 3, DC = 3, DC = 3, DC = 3, DC = 3, DC = 3, DC = 2, ADC / AD, sin, ADC = 2 √ 3-2 ∵ BD = DC / 2 ∵ BD = DC / 2
In △ ABC, D is a point on BC, BD is equal to 1 / 2dc, ∠ ABC = 120 ° ad = 2, if △ ABC area = 3-radical 3, then ∠ BAC=
Let BC = 3x
S △ ABC = 3-radical 3 = 1 / 2sin120 °× ab × 3x
cos120°=(x^2+AB^2-AD^2)/2×x×AB
The length of AB can be obtained by the combination of the above two formulas, and all sides can be obtained. If ABC is a solvable triangle, then ∠ BAC can be obtained
In △ ABC, ∠ 90 ° D is the point on BC and BD = 100, ∠ ADC = 60 ° SINB = radical 2 / 2, find the length of AC ∠C=90°
Because ∠ C = 90 ° SINB = radical 2 / 2; so △ ABC is a right triangle (draw)
If AC is x, then CD is X-100
So CD: AC = 1: radical 3
Root number: 3 x: 1
X = (100 root number 3) / (root number 3-1) = 150 + 50 root number 3
As shown in the figure, △ ABC, ab = AC = 2, BC = 2 3. If the point D is on the edge of BC, ∠ ADC = 45 ° then what is the length of ad?
A vertical line is drawn from a to BC, and the vertical foot is e,
∵AB=AC，BC=2
3，
∴BE=
3，
∵AB=2
∴cosB=BE
AB=
Three
Two
∴B=30°
∴AE=BE•tan30°=1
∵∠ADC=45°
∴AD=AE
sin∠ADC=
2．
As shown in the figure, in △ ABC, the vertical bisector of AB = AC and AC intersects AC at point D and ab at point E. if AE = BC, is point e the golden section point of segment AB? Please tell us your reasons
I think there are three situations:
1. When AB is greater than BC, e is the golden section point of line ab
2. When AB = AC = BC, e is the golden section point of line AB, which is not true. At this time, e and B are the same point
3. When AB is less than BC, I didn't calculate this. It should be: B is the golden section of AE
As shown in the figure, in △ ABC, the vertical bisector of AB intersects AB at point D and AC at point e. it is proved that AE = 2ce
Connect be,
∵ in △ ABC, ∵ C = 90 °, a = 30 °,
∴∠ABC=90°-∠A=60°，
∵ De is the vertical bisector of ab,
∴AE=BE，
∴∠ABE=∠A=30°，
∴∠CBE=∠ABC-∠ABE=30°，
In RT △ BCE, be = 2ce,
∴AE=2CE．
The edge of triangle ABC is ab = 8, AC = 4, the bisector of angle a intersects with the vertical bisector of BC at point D. the straight line passing through point D is de ⊥ AB, DF ⊥ AC, and the perpendicular foot is e, f to find AE
Company, DB, DC
∵ the bisector of angle a intersects with the vertical bisector of BC at point D. the straight line passing through point D is de ⊥ AB, DF ⊥ AC, and the perpendicular foot is e
∴DB=DC,DE=DF
And ? DEB = ∠ DFC = 90
∴ΔDEB≌ΔDFC(HL)
∴BE=CF
AE = AF can be obtained by using Δ ade ≌ Δ ADF
AE=AB-BE=8-BE=AF=AC+CF=4+CF
∴BE=CF=2
∴AE=6
In the triangle ABC, BC is equal to 10, the vertical bisector of AB and the vertical bisector of AC intersect BC at points D and e respectively, and De is equal to 4. Find the length of AD + AE If there is no picture, there will be no question. It seems that there are two answers
Analysis: draw a graph, according to the equal distance between the points on the vertical bisector and the two ends of the line segment, we can get ad = BD, AE = CE, and then discuss the solution in two cases
∵ the vertical bisectors of AB and AC intersect BC at points D and e respectively,
∴AD=BD,AE=CE,
∴AD+AE=BD+CE,
∵BC=10,DE=4,
As shown in Figure 1, AD + AE = BD + CE = bc-de = 10-4 = 6,
As shown in Figure 2, AD + AE = BD + CE = BC + de = 10 + 4 = 14,
In conclusion, AD + AE = 6 or 14
So the answer is: 6 or 14
In the triangle ABC, the angle B is equal to 22.5 degrees, the vertical line of AB intersects BC on D, and DF is perpendicular to AC and F, and intersects with the high AE on BC side at g. it is proved that EG is equal to E Urgent! Process required
The proof should be eg = EC, prove: connect ad, from the meaning of the question: ? B = ∵ bad = 22.5 degrees; BDA = 135 degrees ade = 45 degrees and ? AE ⊥ DC ⊥ AED is an isosceles right triangle ? AE = de and ? EGF + EGD = 180 degrees and ∵ in quadrilateral egfc, ? GEC = ∠ GFC = 90 degrees ? C + EGF = 180 DGE
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