(1 / 2) we know that the two high BD and CE of the acute triangle ABC intersect at O, and ob = OC, and prove that: (1) the triangle ABC is isosceles triangle (1 / 2) it is known that the two high BD and CE of the acute triangle ABC intersect at the point O, and ob = OC. It is proved that: (1) the triangle ABC is an isosceles triangle. (2) whether o is at the angular plane of the angle ABC is determined

(1 / 2) we know that the two high BD and CE of the acute triangle ABC intersect at O, and ob = OC, and prove that: (1) the triangle ABC is isosceles triangle (1 / 2) it is known that the two high BD and CE of the acute triangle ABC intersect at the point O, and ob = OC. It is proved that: (1) the triangle ABC is an isosceles triangle. (2) whether o is at the angular plane of the angle ABC is determined

(1) The two high BD and CE of ∵ ob = OC, ∵ OBC = ∵ OCB, ∵ acute angle △ ABC intersect at O, ∵ BEC = ∵ BDC = 90 °, ∵ BEC + ∵ BCE + ∵ ABC = ∵ BDC + ∵ DBC + ∵ ACB = 180 °, ∵ ABC = ∵ ACB, ∵ AB = AC, ∵ ABC is isosceles triangle; (2) connect AO and extend BC to F, ∵ AB = AC

As shown in the figure, the high BD and CE of △ ABC intersect at O, and ob = OC. Are AB and AC equal? Why?

The reasons are as follows: ∵ ob = OC, ∵ OBC = ∠ OCB. ∵ CE ⊥ AB, BD ⊥ AC, ∵ BEC = ∠ CDB = 90 °, ∵ BC = BC, ≌ BCE ≌ CBD. ∵ EBC = ∠ DCB. ≌ AB = AC

As shown in the figure, the high BD and CE of △ ABC intersect at O, and ob = OC. Are AB and AC equal? Why?

The reasons are as follows: ∵ ob = OC, ∵ OBC = ∠ OCB. ∵ CE ⊥ AB, BD ⊥ AC, ∵ BEC = ∠ CDB = 90 °, ∵ BC = BC, ≌ BCE ≌ CBD. ∵ EBC = ∠ DCB. ≌ AB = AC

In △ ABC, ab = AC, BD and CE are the heights of triangles, BD and CE intersect at point O. note: OB = OC

In the triangle boh and triangle COH (SAS) BH = HC (isosceles triangle three lines in one) ∠ AHB = ∠ AHC (vertical 90 ° angle) Oh = Oh (common edge) ｜ ob = OC method 2: Triangle EOB congruent triangle doc ∠ BeO = ∠ CDO (vertical 90 ° angle) eo = do (angle bisection)

A rectangular iron plate is 6 meters long and 2 / 3 of its length wide. How many square meters is the area and perimeter of this steel plate?

Width = 6 * 2 / 3 = 4m
Area = 6 * 4 = 24 square meters
Perimeter = 2 * (6 + 4) = 20m

A rectangular iron plate. 15m in length and 2 / 3 in width. Cut the largest circle on the plate. What is the area and perimeter of the circle?
If you cut off the largest semicircle, what are the perimeter and area of the semicircle?

Width = 15 × 2 / 3 = 10m
Area = 10 × 10 × 3.14 = 314 square meters
Perimeter = 10 × 2 × 3.14 = 62.8m
Semicircle
Perimeter = 15 × 3.14 × 1 / 2 + 15 = 38.55m
Area = 15 × 15 × 3.14 × 1 / 2 = 353.25 square meters

A rectangular iron plate is 6 meters long, 13 meters wide and 13 meters long. What's the area of this iron plate? What's the perimeter?

6 × 13 = 2 (meters), 6 × 2 = 12 (square meters); (6 + 2) × 2 = 16 (meters); a: the area of this iron plate is 12 square meters, and the perimeter is 16 meters

The length of a rectangular iron plate is 4 meters, and the width ratio is 1 / 2 of the length. How many square meters is the area of this iron plate? How many meters is the perimeter?

Width = 4 × (1-2 / 2) = 2m
Area = 4 × 2 = 8 square meters
Perimeter = 2 × (4 + 2) = 12m

A circular iron plate and a square iron plate, their edge circumference is 37.68 cm, which iron plate area is large? How much is each?

Square iron plate area: (37.68 △ 4) × (37.68 △ 4), = 9.42 × 9.42, = 88.7364 (square centimeter); circular iron plate area: 3.14 × (37.68 △ 3.14 △ 2) 2, = 3.14 × 36, = 113.04 (square centimeter), 113.04 ＞ 88.7364; answer: Circular iron plate area is large, square iron plate area is 88.7364 square centimeter, circular iron plate area is 113.04 square centimeter

There is a semicircular iron plate with a radius of 2 cm. What is the circumference of this iron plate?

Circumference of circle = 2 π R  circumference of semicircle = π r circumference of iron plate = circumference of semicircle + diameter length = π R + 4 = 2 π + 4