As shown in the figure, in △ ABC, ab = AC, AE ⊥ AB in a, ∠ BAC = 120 ° AE = 3cm

As shown in the figure, in △ ABC, ab = AC, AE ⊥ AB in a, ∠ BAC = 120 ° AE = 3cm

Through point a, AF ⊥ BC intersects BC with F, ∵ AB = AC, ∠ BAC = 120 °, ∵ B = ∠ C = 30 °, BC = 2BF. In RT △ BAE, ab = AE · cot 30 ° = 3 × 3 = 33, in RT △ AFB, BF = ab · cos 30 ° = 33 × 32 = 92, ∵ BC = 2BF = 2 × 92 = 9

As shown in the figure, in △ ABC, ab = AC, AE ⊥ AB in a, ∠ BAC = 120 ° AE = 3cm

Through point a, AF ⊥ BC intersects BC with F, ∵ AB = AC, ∠ BAC = 120 °, ∵ B = ∠ C = 30 °, BC = 2BF. In RT △ BAE, ab = AE · cot 30 ° = 3 × 3 = 33, in RT △ AFB, BF = ab · cos 30 ° = 33 × 32 = 92, ∵ BC = 2BF = 2 × 92 = 9

In ABC, ab = 7, BC = 6, AC = 4, ad and AE are respectively the middle line and high line on the side of BC, and the length of De is calculated

In △ ABC, COSC = 1 / 16 is obtained by cosine theorem
In △ ace, COSC = CE / AC = 1 / 16
CE = 1 / 4
DE=CD-CE=3-1/4=2.75

In the triangle ABC, it is known that ad AE is the height and the median line AB = 9, AC = 5, BC = 8 on the edge of BC respectively, and the length of De is obtained

In the right triangle ADB and ADC, according to the Pythagorean theorem, 81-bd = 25 - (8-bd), BD = 7.5 and be = 4, so de = 3.5