As shown in the figure, in △ ABC, ab = AC, ad is the angular bisector of ∠ BAC, ad = 8cm, BC = 6cm, and points E and F are two points on ad, then the area of shadow part in the figure is () A. 48 B. 24 C. 12 D. 6

As shown in the figure, in △ ABC, ab = AC, ad is the angular bisector of ∠ BAC, ad = 8cm, BC = 6cm, and points E and F are two points on ad, then the area of shadow part in the figure is () A. 48 B. 24 C. 12 D. 6

∵ AB = AC, ad is the bisector of ∵ BAC,
∴BD=DC=8，AD⊥BC，
The △ ABC is symmetric about the straight line ad,
﹣ B and C are symmetrical about the straight line ad,
The △ CEF and △ bef are symmetric with respect to the straight line ad,
∴S△BEF=S△CEF，
The area of ABC is 1
2×BC×AD=1
2×8×6=24，
The area of the shaded part in the figure is 1
2S△ABC=12．
Therefore, C

As shown in the figure, in triangle ABC, ab = 2cm, BC = 4cm, what is the ratio of height ad to ce of triangle ABC?

It's the area of the triangle
AB*CE=AD*BC
It turns out to be 1:2

As shown in the figure, ad and CE are two heights of △ ABC, known as ad = 10, CE = 9, ab = 12 (1) Find the area of △ ABC; (2) Find the length of BC

（1）∵CE=9，AB=12，
The area of △ ABC = 1
2×12×9=54；
(2) Area of △ ABC = 1
2BC•AD=54，
That is 1
2BC•10=54，
BC = 54
5．

As shown in the figure, it is known that in △ ABC, ad is high, CE is midline, DC = be, DG ⊥ CE, G is perpendicular foot It is proved that: (1) g is the midpoint of CE; (2) B = 2 ∠ BCE

It is proved that: (1) connect De;
∵ ad ⊥ BC, e is the midpoint of ab,
﹤ De is the center line on the oblique side of RT △ abd, i.e., de = be = 1
2AB；
∴DC=DE=BE；
And ∵ DG = DG,
∴Rt△EDG≌Rt△CDG；（HL）
∴GE=CG，
/ / G is the midpoint of CE
(2) According to (1), be = de = CD;
∴∠B=∠BDE，∠DEC=∠DCE；
∴∠B=∠BDE=2∠BCE．

As shown in the figure, △ ABC, the lengths of high AD and CE are 2cm and 4cm respectively. What is the ratio of AB to BC?

∵S△ABC=1
2AB•CE=1
2BC · ad, height ad = 2cm, CE = 4cm,
∴AB•CE=BC•AD
∴AB
BC＝AD
CE＝2
4＝1
2．

As shown in the figure, in the triangle ABC, ab = 2, AC = 3, BC = 4, ad, BF, CE are the three heights of the triangle ABC. Calculate the ratio of the three heights ad, BF, CE

Using the same area, 1 / 2 * BF * AC = 1 / 2 * BC * ad = 1 / 2 * AB * ce is obtained
BF = 2 / 3ce, CE = 2ad
So ad: BF: CE = 3:4:6

In the triangle ABC, ad is the height on the BC side, ab = 15cm, AC = 13cm, ad = 12cm. Find the area of the triangle

BC=√(15^2-12^2)+√(13^2-12^2)
=9+5=14
S=(1/2)*BC*AD=14*12/2=84
☆⌒_ I hope I can help you~

In the triangle ABC, if AB = 15, AC = 20, and the height ad on BC side is 12, then the area of triangle ABC is

∵AB=15,AC=20,AD⊥BC
∴BD²=AB²-AD² DC²=AC²-AD²
∴BD=9 DC=16 BC=BD+DC=25
S△ABC=25X12X1/2=150

In the triangle ABC, if AB = 15, AC = 13, height ad = 12, then the area of triangle ABC is

Ad is high, and there is one with Pythagorean theorem
AB^2=BD^2+AD^2,15^2=BD^2+12^2,BD=9
AC^2=CD^2+AD^2,13^2=CD^2+12^2,CD=sqrt(13)
So area = 1 / 2 * BC * ad = 1 / 2 * (BD + CD) * ad = 54 + 6sqrt (13)

In the triangle ABC, ab = 5cm, AC = 13cm, and the center line ad = 6cm on BC side, calculate the area of triangle ABC

Triangle ABC area = triangle ABA 'area triangle ABA' area = √ [P (P - a) (P - b) (P - C)] (Helen's formula)) (P = (a + B + C) / 2) P = (AB + a 'B + AA') / 2) = (13 + 5 + 12) / 2 = 15s = √ [P (P (P & nbs)