Given that the circumference of the isosceles triangle ABC is 40cm, ad is the height on the bottom edge, and the circumference of the triangle abd is 30cm, what is the length of ad

Given that the circumference of the isosceles triangle ABC is 40cm, ad is the height on the bottom edge, and the circumference of the triangle abd is 30cm, what is the length of ad

Let AB = AC = X
BC=40-2X
BD=1/2 BC=20-X
The circumference of abd is 30 cm
AD=30-AB-BD
=30-X-(20-X)
=10

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

As shown in the figure, if the bottom edge BC of isosceles △ ABC is 16, and the height ad on the bottom edge is 6, then the length of waist length AB is______ .

∵ isosceles ᙽ ABC's bottom edge BC is 16, and the height ad on the bottom edge is 6,
∴BD=8,AB=
AD2+BD2=
62+82=10.

As shown in the figure, if the bottom edge BC of isosceles △ ABC is 16, and the height ad on the bottom edge is 6, then the length of waist length AB is______ .

∵ isosceles ᙽ ABC's bottom edge BC is 16, and the height ad on the bottom edge is 6,
∴BD=8,AB=
AD2+BD2=
62+82=10.

If in the isosceles triangle ABC, the waist length AB = 13cm, and the ad on the bottom edge = 12cm, then the bottom edge BC =? I always think it should be the height of the bottom edge ad = 12cm If there is no problem with this problem, BC = 10cm Is there any problem with this problem

If there is no isosceles triangle,
BC=2*BD
BD*BD=13*13-12*12
BC=10
If you are satisfied with the answer, please set it for me to adopt,

Cut the isosceles triangle paper ABC along the height ad on the bottom BC into two triangles. How many kinds of parallelograms can you spell out with these two triangles? Find out their diagonal lengths respectively

(1) The diagonal length is m
(2) The diagonal length is n and [(2 * h) ^ 2 + n ^ 2]
(3) The length of diagonal is m and [(2 * n) ^ 2 + H ^ 2]

As shown in figure a, in △ ABC, ab = AC, and there is any point P on the bottom BC, then the sum of the distances from point P to two waists is equal to the fixed length (the height on the waist), that is, PD + PE = cf. if the point P is on the extension line of BC, what kind of equality relationship exists between PD, PE and CF? Write your guess and prove it

My guess is: the relationship between PD, PE and CF is PD = PE + CF

As shown in figure a, in △ ABC, ab = AC, and there is any point P on the bottom BC, then the sum of the distances from point P to two waists is equal to the fixed length (the height on the waist), that is, PD + PE = cf. if the point P is on the extension line of BC, what kind of equality relationship exists between PD, PE and CF? Write your guess and prove it

My guess is that the relationship among PD, PE and CF is PD = PE + CF
If AP is connected, s △ PAC + s △ cab = s △ PAB,
∵S△PAB=1
2AB•PD,S△PAC=1
2AC•PE,S△CAB=1
2AB•CF,
And ∵ AB = AC,
∴S△PAC=1
2AB•PE,
∴1
2AB•PD=1
2AB•CF+1
2AB•PE,
That is 1
2AB(PE+CF)=1
2AB•PD,
∴PD=PE+CF.

As shown in the isosceles △ ABC, ab = AC, the sum of the distances from any point P on the bottom edge BC to the two waist is equal to the height on one waist Can you prove this conclusion by area method?

I gave you the email of 12345

As shown in the figure, in the triangle ABC, the angle B = 90 degrees, ab = 3, BC = 4, AC = 5, (1) is there a point P in the triangle ABC whose distance from each side is equal If so, please make this point and explain the reason (2) Find the distance

The distance from the center of a triangle (the intersection of the bisectors) to the three sides is equal