As shown in the figure: the circumference of △ ABC is 24cm, ab = 10cm, the vertical bisector De of edge AB intersects BC, the edge is at point E, and the perpendicular foot is D, calculate the circumference of △ AEC

As shown in the figure: the circumference of △ ABC is 24cm, ab = 10cm, the vertical bisector De of edge AB intersects BC, the edge is at point E, and the perpendicular foot is D, calculate the circumference of △ AEC

The vertical bisection is ab
 be = AE (2 points)
The circumference of △ ace = AE + EC + AC = be + CE + AC = BC + AC
The circumference of ABC is 24cm, ab = 10cm
∴BC+AC=24-10=14cm
The circumference of ACE is 14 cm

In the triangle ABC, ab = AC, the vertical bisector De of AB intersects BC with E, and the perpendicular foot is d. if the circumference of triangle ABC and AEC are 26 cm and 18 cm respectively, the vertical bisector of AB is 0 Find the length of each side of the triangle ABC

AB + BC + Ca = 26 cm AE + EC + Ca = 18 because AE + EC = be + EC = BC, so BC + AC = 18, ab = 8 cm, because AB = AC, so AC = 8 cm, so BC = 10 cm, so triangle

As shown in the figure, in △ ABC, if the vertical bisector of AC intersects AC at e and BC at D, the circumference of △ abd is 12cm, AC = 5cm, then AB + BD + ad=______ cm;AB+BD+DC=______ Cm; the circumference of △ ABC is______ cm.

∵ De is the vertical bisector of line AC,
∴AD=CD,
∴AD+BD=CD+BD,
The circumference of abd is 12 cm,
∴AB+BD+AD=12cm,AB+BD+DC=12cm,
∵AC=5cm,
The circumference of △ ABC = (AB + BD + DC) + AC = 12 + 5 = 17cm
So the answer is: 12, 12, 17

As shown in the figure, the circumference of △ ABC is 19cm, the vertical bisector De of AC intersects BC at D, e is perpendicular foot, AE = 3cm, then the circumference of △ abd is______ cm.

∵ the vertical bisector De of AC intersects BC with D, and E is the perpendicular foot
∴AD=DC,AC=2AE=6cm,
The circumference of ABC is 19 cm,
∴AB+BC=13cm
The circumference of △ abd = AB + BD + ad = AB + BD + CD = AB + BC = 13cm
Therefore, fill in 13

In △ ABC, ab = AC, the vertical bisector Mn of AB intersects AC at point D and ab at point E. if AE = 6 and the circumference of △ CBD is 20, calculate the circumference of △ ABC

∵ Mn vertical bisection ab
∴AD=BD
AE=BE=1/2AB=6,AB=12
∵ △ CBD perimeter = BC + BD + CD = BC + CD + ad = BC + AC = 20
∴△ABC=AB+BC+AC=12+20=32

In △ ABC, De is the vertical bisector of AC, AE = 5cm, and the circumference of △ CBD is 24cm. Calculate the circumference of △ ABC

In ABC, De is the vertical bisector of AC,
∴AD=CD,CE=AE=5cm,
∴AC=AE+CE=10cm,
The circumference of CBD is 24cm,
∴BC+CD+BD=BC+AD+BD=BC+AB=24(cm),
The circumference of △ ABC is: AC + AB + BC = 10 + 24 = 34 (CM)

As shown in the figure, in △ ABC, De is the vertical bisector of AC, AE = 5cm, the circumference of △ ABC is 25cm, and calculate the circumference of △ abd

The circumference of △ ABC = AB + AC + BC
=AB+2AE+BD+DC=25
De bisects AC vertically, so ad = DC
The circumference of △ abd = AB + BD + ad
=AB+BC=25-AC=25-10=15Cm

As shown in the figure, given that AB is 2cm longer than AC, the vertical bisector of BC intersects AB at D, crosses BC at e, and the circumference of △ ACD is 14cm, then ab=______ cm,AC=______ cm.

∵ de bisects BC vertically,
∴DB=DC.
∵AC+AD+DC=14cm,
∴AC+AD+BD=14cm,
AC + AB = 14cm
And ∵ AB-AC = 2cm,
Let AB = xcm, AC = YCM
According to the meaning of the title, we get
x+y=14
x−y=2 ,
The solution
x=8
y=6
The length of AB is 8cm and that of AC is 6cm

As shown in the figure, in △ ABC, ab = 5cm, AC = 3cm, and the vertical bisector of BC intersects AB, BC at D and e respectively, then the circumference of △ ACD is______ cm.

∵ De is the vertical bisector of BC,
∴CD=BD,
The circumference of △ ACD = AC + CD + ad = AC + AD + BD = AC + AB,
AC = 3cm, ab = 5cm,
The circumference of △ ACD is 3 + 5 = 8cm
So the answer is: 8

As shown in the figure, it is known that in trapezoidal ABCD, BD bisection ∠ ABC, ∠ a = 120 ° and BD = BC = 4 3, then s trapezoid ABCD = () A. 4 Three B. 12 C. 4 3-12 D. 4 3+12

As shown in the figure: AF ⊥ BD at point F, de ⊥ BC at point e through point D,
∵ a = 120 ° in trapezoidal ABCD,
∴∠ABC=60°,∠ADB=∠DBC,
∵ BD bisection ∵ ABC,
∴∠ABD=∠DBC=30°,
∴∠ABD=∠ADB=30°,
∴AB=AD,
∵BD=BC=4
3,
∴BF=2
3,
∴AB=AD=BF
cos30°=4,
De = 1
2BD=2
3,
Therefore, s trapezoid ABCD = 1
2(AD+BC)×DE=1
2×(4+4
3)×2
3=4
3+12.
Therefore, D