It is known that Mn is the vertical bisector of line AB, and C and D are two points on Mn (1) Δ ABC, △ abd are isosceles triangles; （2）∠CAD=∠CBD．

It is known that Mn is the vertical bisector of line AB, and C and D are two points on Mn (1) Δ ABC, △ abd are isosceles triangles; （2）∠CAD=∠CBD．

It is proved that: (1) ∵ Mn is the vertical bisector of line AB, C and D are two points on Mn
∴AC=BC，AD=BD，
That is, △ ABC, △ abd is isosceles triangle;
（2）∵AC=BC，AD=BD，
∴∠CAB=∠CBA，∠DAB=∠DBA，
∴∠CAB+∠DAB=∠CBA+∠DBA，
That is ∠ CAD = ∠ CBD

As shown in the figure, △ ABC, ab = AC = 14cm, D is the midpoint of AB, de ⊥ AB intersects AC with e at D, and the circumference of ⊥ EBC is 24cm. Find the length of BC

In △ Abe,
∵ D is the midpoint of AB, de ⊥ AB is at D and AC is at E,
∴AE=BE；
In △ ABC,
∵AB=AC=14cm，AC=AE+EC，
And ∵ CE + be + BC = 24cm,
∴BC=10cm．

As shown in the figure, in the triangle ABC, ab = AC, angle a = 40 degrees, the vertical bisector Mn of AB intersects AC at point D. find the degree of angle DBC

∵ Mn is the vertical bisector of ab
﹤ abd is an isosceles triangle, i.e. ﹤ DBA = ∠ DAB = 40 °
And ∵ AB = AC, ∵ a = 40 ᙽ
∴∠ABC=70°
∴ ∠DBC=∠ABC-∠ADB=30°

As shown in the figure, in △ ABC, ab = AC, ∠ a = 80 °, the vertical bisector of AB intersects the extension line of AC and the point D, and calculate the degree of ∠ DBC

In the case of ∵ AB = AC, ∵ AB = AC, ∵ a = a = 80 ∵ a = 80 ∵ a ∵ a = 80 ∵ a ∵ a = a = 80 ∵ a ∵ a ∵ am = BM ⊙ ab ⊙ MD

As shown in the figure, ab = AC in △ ABC, the vertical bisector Mn of AB intersects AC at point D. if AC + BC = 10cm, the circumference of △ DBC is______ .

∵ Mn vertical bisection AB,
∴DA=DB．
The circumference of △ DBC = BC + BD + DC
=BC+DA+DC=BC+AC=10cm．
So the answer is: 10cm

In the triangle ABC, ab = AC, the vertical bisector Mn of AB intersects AC at D. if AB = 10, BC = 6, calculate the circumference of the triangle DBC

Ah, since Mn is the vertical bisector of AB, we can see that ad = BD, then the circumference of DBC = BC + BD + DC = BC + AD + DC = BC + AC = 6 + AC = 6 + AB = 6 + 10 = 16

As shown in the figure, ab = AC, ∠ a = 40 °, the vertical bisector Mn of AB intersects AC at point D, and the degree of ∠ DBC is calculated

∵AB=AC，
∴∠ABC=∠ACB=180°−∠A
2＝180°−40°
2=70°，
∵ the vertical bisection ab of Mn,
∴DA=DB，
∴∠A=∠ABD=40°，
∴∠DBC=∠ABC-∠ABD=70°-40°=30°．
So the answer is: 30 degrees

As shown in the figure, in △ ABC, ∠ a = 50 °, ab = AC, and the vertical bisector De of AB intersects AC with D, then the degree of ∠ DBC is () A. 15° B. 20° C. 30° D. 25°

The solution is known, ∠ a = 50 °, ab = AC ∠ ABC = ∠ ACB = 65 °
And ∵ De is vertical and bisection ab ∵ DB = ad
∴∠ABD=∠A=50°
∴∠DBC=∠ABC-∠ABD=65°-50°=15°．
Therefore, a

As shown in the figure, in the triangle ABC, ab = AC, angle a = 40 degrees, the vertical bisector Mn of AB intersects AC at point D, and calculate the degree of angle DCB

Angle DCB = 70 degrees,
You should want to know the degree of the angle DBC,
Connect to DB,
If AB = AC, angle a = 40 degrees, then: angle B = angle c = 70
Then the vertical bisection of AC by Mn can be proved; Da = dB, angle a = angle DBA = 40
Then: angle BDC = 80, and angle c = 70
Then: angle DBC = 30

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