In the triangle ABC, ab = AC, the center line be of triangle ABC divides the circumference of triangle into two parts of 9cm and 12cm, and calculates the length of BC in triangle ABC You have to draw your own pictures, The length of the center line is also included in the perimeter of the two parts

In the triangle ABC, ab = AC, the center line be of triangle ABC divides the circumference of triangle into two parts of 9cm and 12cm, and calculates the length of BC in triangle ABC You have to draw your own pictures, The length of the center line is also included in the perimeter of the two parts

If the triangle AB + AE = 9, then CE + BC = 12, because AE = EC, ab = AC, therefore: 2Ab + BC = 12 + 9 = 21bc-ab = 12-9 = 3 can be obtained from above: 3bC = 27BC = 9. Similarly, if AE + AB = 12, EC + BC = 9, 2Ab

In the triangle ABC, ab = AC, CD is the height on the bottom edge, the circumference of triangle ABC is 16cm, and the circumference of triangle ACD is 12cm. Find the length of AD

AB = AC, ad is the height on the bottom edge, BD = CD
The circumference of triangle ABC is 16cm, that is ab + AC + BC = 16, that is 2Ac + 2CD = 16, AC + CD = 8
The circumference of triangle ACD is 12cm, AC + CD + ad = 12, ad = 12-8 = 4

As shown in the figure, we know that the base edge of the isosceles triangle ABC is BC = 20cm, D is a point on waist AB, and CD = 16cm, BD = 12cm. Find the circumference of △ ABC

∵ BC = 20cm, CD = 16cm, BD = 12cm ᙽ BC  BC  BC  BC  BC ? BC ? BC ? AB = AC ? AC = AD + BD = AD + 12 ? AC ? AC ? ad ? ad ? ad ? ad ? ad ? ad ? BC

Known isosceles triangle ABC, ab = AC, BC = 20cm, D is a point on AB, and CD = 16cm, BD = 12cm

Solution: let ad be X
Because DB = 12 BC = 20 CD = 16
And this is a group of Pythagorean numbers
So the triangle DBC is a right triangle
The square of AC = the square of X + the square of 16
So (x + 12) 2 = x2 + 16 squared
X=14/3
AB+AC+BC=160/3

Given the isosceles triangle ABC, base BC = 20, D is a point on AB, and CD = 16, BD = 12, find the length of AD

In △ BCD, △ BCD is a right triangle from 122 + 162 = 202. Let ad = x, then AC = 12 + X,
According to Pythagorean theorem, x 2 + 162 = (x + 12) 2 and x = 14 are obtained
3.
∴AD=14
3.

Given that the median line of the waist length of an isosceles triangle divides its circumference into two parts of 15cm and 12cm, the base length of this isosceles triangle is___ .

∵ the circumference of an isosceles triangle is 15cm + 12cm = 27cm. If the waist length and bottom edge length of the isosceles triangle are xcm and YCM respectively, we can get  x + 12x = 1512x + y = 12 or x + 12x = 1212x + y = 15 (4 points), x = 10Y = 7 or x = 8y = 11 (3 points)

As shown in the figure, in the isosceles triangle ABC, AC = AB = 12cm, the vertical bisector of AB intersects AC at point D, and the circumference of △ BCD is 21cm, what is the length of bottom BC

∵ the vertical bisector of AB intersects AC at point D
∴AD=BD
The circumference of △ BCD
=BD+BC+CD
=AD+CD+BC
=AC+BC
∵AC+BC=21
AC=12
∴BC=21-12=9

It is known that in the isosceles triangle ABC, ab = AC and the circumference is 16 cm. The median line BD on the edge of AC divides △ ABC into two triangles with perimeter difference of 4cm Find the length of each side of the triangle ABC

In this triangle, let AB = 2x, ad = AC = x, BC = y
4X+Y=16
2X-Y=4
These two triangles are actually the difference between AB and BC. BD is the common line
Hope to adopt

It is known that the circumference of the isosceles triangle ABC is 16cm, ad is the bisector of the vertex angle, AB ratio ad = 5:4, and the circumference of △ abd is 12cm

Ad is the top bisector and the triangle is isosceles triangle
Ad is perpendicular to BC and BD = BC
AB / ad = 5:4
That is cos ∠ bad = 4 / 5
Sin∠BAD=3/5
AB:BC=5:6
2AB+BC=16
AB=AC=5
BC=6

As shown in the figure, the circumference of the isosceles △ ABC is 50cm, ad is the height on the bottom edge, and the circumference of △ abd is 40cm, then the length of ad is______ cm.

∵ ad is the height on the bottom edge,
∴BD=CD,
The circumference of ABC is 50cm,
∴AB+BD=1
2×50=25cm,
The circumference of abd is 40 cm,
∴AD=40-25=15cm.
So the answer is: 15