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It is known that ad is the median line of the triangle ABC, e is the extension of CE at any point on ad, and the intersection of AB and F is proved to be AE / ad = 2AF / BF
Proved AE: de = 2AF: BF
When DH ∥ AB is crossed with CF at point D, then △ DHE ∷ AFE, so AE: de = AF: DH
∵BD=CD,DH‖AB
∴DH=1/2BF
∴AE:DE=AF:1/2BF
AE: de = 2AF: BF
As shown in the figure, △ ABC, D is a point on BC. If AB = 10, BD = 6, ad = 8, AC = 17, calculate the area of △ ABC
∵BD2+AD2=62+82=102=AB2，
ν Δ abd is a right triangle,
∴AD⊥BC，
In RT △ ACD, CD =
AC2−AD2＝
172−82＝15，
∴S△ABC=1
2BC•AD＝1
2(BD+CD)•AD＝1
2×21×8＝84，
So the area of △ ABC is 84
A: the area of △ ABC is 84
In the triangle ABC, ab = 17, AC = 15, and the median line ad = 4 on the BC side, calculate the area of the triangle ABC
Extend ad to point e so that de = ad, connect be, CE
Then the quadrilateral ABEC is a parallelogram
∴BE=AC=15
∵AD=4
∴AE=8
∵8²+15²=17²
∴∠AEB=90°
The area of ABEC of parallelogram is 15 * 8 = 120
The area of △ ABC = 60
Given the triangle ABC, ab = 10, BC = 9, AC = 17, find the height ad on the edge of BC
In this paper, we obtain the obtuse angle of the coppery = 17-17 * S-21, and let the square of the cosine = 17 * y = 17-21, and let the square of the cosine = 17 * y = 17-21, and let the square of the cosine = 17 * x-21 be the square of the cosine
Known: in the triangle ABC, ab = 25, AC = 17, ad is the height of BC, and ad = 15, find the length of BC side
Because ad is higher than BC,
So triangle abd and triangle ACD are right triangles,
So AB squared = ad squared + BD squared, AC squared = ad squared + CD squared,
Because AB = 25, AC = 17, ad = 15,
So BD squared = 400, CD squared = 64,
BD=20,CD=8,
So BC = BD + CD = 20 + 8 = 28
Given the triangle ABC, ab = 10, BC = 21, AC = 17, find the height of BC edge?
As ad ⊥ BC, let BD = X
From the meaning of the title, we can get
10²-x²=17²-（21-x）²
100-x²=289-441+42x-x²
252=42x
X=6
∴AD=√(10²-6²)=8
A: the height on the side of BC is 8
In the triangle ABC, AB is equal to AC equal to 17, BC to 16. Find the area of triangle ABC and the height of waist AC
As shown in the figure, in △ ABC, ab = ad, DC = BD, de ⊥ BC, de intersects AC at point E, be intersects ad at point F The results showed that: (1) BDF △ CBA; (2) AF = DF
It is proved that: (1) BD = DC, de ⊥ BC, EB = EC. EBD = ∵ C. (3 points)
As shown in the figure, in the triangle ABC, BD = DC = AC, e is the midpoint of DC, and the bisection angle BAE of ad is proved
As shown in the figure, extend AE to F, make ef = AE, and connect DF
In △ ace and △ FDE,
AE=EF,∠AEC=∠DEF,CE=DE
∴△ACE≌△FDE（SAS）
∴DF=AC=BD,∠F=∠FAC,∠C=∠FDC
∵AC=CD
∴∠CAD=∠ADC
∵∠ADB=∠C+∠CAD=∠FDC+∠ADC=ADF
In △ abd and △ AFD
AD=AD,∠ADB=∠ADF,BD=DF
∴△ABD≌△AFD（SAS）
∴∠BAD=∠FAD,
That is, ad bisection angle BAE
As shown in the figure, in the isosceles triangle ABC, ab = AC, ∠ a = 20 °, D is a point on the edge of AB, ad = BC, connecting CD, then the size of ∠ BDC is 3030 °
Take AC as an edge, make an equilateral triangle △ ace on the outside of △ ABC, connect de. ∵ AB = AC, vertex angle ∵ a = 20 °, ABC = 80 °, ∵△ ace is an equilateral triangle, AC = AE = CE, EAC = 60 °, ead = 80 °, AE = AB, ∵ ad = BC,
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