As shown in the figure, in the isosceles △ ABC, ab = AC. the semicircle with diameter AB intersects BC at point D and AC at point E De = 40 °, find ∠ A and Degree of AE

As shown in the figure, in the isosceles △ ABC, ab = AC. the semicircle with diameter AB intersects BC at point D and AC at point E De = 40 °, find ∠ A and Degree of AE

Connect ad,
∵ AB is the diameter,
∴∠ADB=90°，
Ad ⊥ BC,
∵AB=AC，
∴∠BAD=∠CAD，
Qi
BD=
DE=40°，
∴∠BAD=∠CAD=1
2×40°=20°，
∴∠BAC=40°，
∴∠B=∠C=70°，
Qi
AD=140°，
Qi
AE=
AC-
DE=100°．

Δ ABC is an isosceles triangle, ab = AC, e is a point on AC arc in circle O. the extension line of BC and AE intersects point D and connects CE. It is proved that ab × CE = AE × CD

In △ EBA and △ EDC, it is proved that in △ EBA and △ EDC, ∠ Dec + ∠ CEB + ∠ AEB = 180 ° ? △ ABC is an isosceles triangle ? ABC + ∠ BCA + ∠ cab = 180 °. BCA = ∠ cab ? CEB = ∠ cab (circumference of the same arc) ? ? AEB = ∠ Dec (equivalent substitution) ∠ EDC + ∠ EAB + ∠ ABC = 180 °

As shown in the figure, the vertices a, B, C of △ ABC are all on the circle, and arc AB = arc BC = arc AC, D is a point on arc BC. Connect AD and intercept AE = DC on AD Try to judge the shape of △ BDE and explain the reason

Equilateral triangle

In the RT triangle ABC, ∠ C = 90 ° and arc cut AB with C as the center of the circle. If ad = 4 and BD = 1 are known, then the area of shadow in the graph is

With CD
　　∵ AD＝4、 BD＝1
　　∴ AB = AD + BD = 4 + 1 = 5
∵ the arc centered on C is tangent to point d with ab
⊥ AB (the tangent line of the circle is perpendicular to the radius of the tangent point)
　　∴ ∠BDC = ∠CDA = 90°
　　∵ ∠C＝90°
　　∴ ∠B + ∠A = 90° --------------------------------- ①
　　∵ CD ⊥ AB
　　∴ ∠B + ∠BCD = 90° ------------------------------------ ②
From ① and ②, it is concluded that: ∠ a = ∠ BCD
In RT △ BCD and RT △ CAD
∠ BDC = ∠ CDA = 90 ° (proven)
∠ a = ∠ BCD (proved)
　　∴ Rt△BCD ∽ Rt△CAD
　　∴ BD ：CD = CD ：AD
The square of CD = BD × ad
　　= 1 × 4
　　= 4
　　∴ CD = 2
Then s △ ABC = (1 / 2) × ab × CD
　　= （1/2）× 5 × 2
　　= 5
　　∵ ∠C = 90°
﹣ the area s fan of the sector formed by the arc with C as the center and two right angle sides (BC, AC)
It is equal to one fourth of the area of a circle with C as its center and CD length as its radius
﹤ s Fan = (1 / 4) × (π × CD squared)
= (1 / 4) × (the square of π × 2)
　　= π
﹣ s negative = s △ ABC -- s fan
　　= 5 -- π

As shown in the figure, in the isosceles triangle ABC, ab = AC, with point C as the center of the circle and CB as the radius, the circle intersects with the edge AB at point D. if ad = BC, In the isosceles triangle ABC, ab = AC, with point C as the center of the circle and CB as the radius, intersect with the edge AB at point D. if ad = BC, find the degree of angle a (2) if AC = 13, BC = 10, find the length of the dazzle length BD

It's a

As shown in the figure, in the triangle ABC, the angle c = 90 ° ad bisects ∠ cab, intersects CB at point D, passes through point D as De, and perpendicular to point E (1) It is proved that △ ACD is equal to △ AED (2). If angle B = 30 ° and CD = 1, find the length of BD

(1) It is proved that because De is perpendicular to AB, angle DEA = angle DEB = 90 degrees, because angle c = 90 degrees, so angle DEA = angle c because ad bisects angle cab, angle CAD = angle ead in triangle ACD and triangle AED, angle ACD = angle AED angle CAD = angle eadad, so triangle ACD is equal to triangle AED (AAS)

As shown in the figure, in the triangle ABC, ∠ C = 90 °, AC = BC, ad is the bisector of ∠ a, intersecting CB at point D, if AC = 3, find BD length

Make de vertical AB through D
Because ad is the bisector of ∠ a
So CD = de AC = AE = 3
Because AB = 2 times root number, AC = 3 root number 2
So be = ab-ae = 3 root number 2-3
So BD = 2 times root number be = 6-3 root sign 2

As shown in the figure, ab = AC in triangle ABC, point D is on the extension line of CB. You can explain: the square of ad - square of AB = BD · CD

It's very simple, go through point a and make the vertical line of BC, and the foot of the perpendicular is point E. then: (the following lines without brackets have squares, and those with brackets have no squares, for convenience of writing)
AD-AE=DE.1
AB-AE=BE=CE.2
Type 2-1 yes
AB-AD=CE-DE
(DE) = (cd-ce)
AB-AD=(2CD*CE)-CD=(CD*BC)-CD=(CD*(BC-CD))=(CD*BD)
That is ab squared - ad squared = BD times CD
You wrote it backwards

As shown in the figure, in right angle trapezoid ABCD, ∠ ABC = 90 °, ad ‖ BC, ab = BC, e is the midpoint of AB, CE ⊥ BD (1) Results: be = ad; (2) Verification: AC is the vertical bisector of ED; (3) Is △ DBC an isosceles triangle? And explain the reason

(1) It is proved that ∵ ABC = 90 ° BD ⊥ EC,
∴∠1+∠3=90°，∠2+∠3=90°，
∴∠1=∠2，
In △ bad and △ CBE,
∠2＝∠1
BA＝CB
∠BAD＝∠CBE＝90° ，
∴△BAD≌△CBE（ASA），
∴AD=BE．
(2) It is proved that ∵ e is the midpoint of ab,
∴EB=EA，
∵AD=BE，
∴AE=AD，
∵AD∥BC，
∴∠7=∠ACB=45°，
∵∠6=45°，
∴∠6=∠7，
And ∵ ad = AE,
⊥ De, and EM = DM,
AC is the vertical bisector of ED;
(3) Δ DBC is an isosceles triangle (CD = BD)
The reasons are as follows.
∵ from (2), CD = CE, from (1), CE = BD,
∴CD=BD．
The △ DBC is an isosceles triangle

As shown in the figure, in right angle trapezoid ABCD, ∠ ABC = 90 °, ad ‖ BC, ab = BC, e is the midpoint of AB, CE ⊥ BD (1) Results: be = ad; (2) Verification: AC is the vertical bisector of ED; (3) Is △ DBC an isosceles triangle? And explain the reason

(1) It is proved that: ? ABC = 90 °, BD ⊥ EC,  1 + ≌⊥ 3 = 90 °, 2 + ⊥ 3 = 90 ° and  1 = ∠ 2. In △ bad and △ CBE, ∠ 2 = 1BA = CB ∠ bad = CBE = 90 °,  bad ≌ △ CBE (ASA),  ad = be. (2) it is proved that  e is the midpoint of AB,  EB = EA, M ad = be