As shown in the figure, in the acute triangle ABC, AB > AC, ad is perpendicular to D, and graph o with AD as diameter intersects AB, AC with E and f respectively

As shown in the figure, in the acute triangle ABC, AB > AC, ad is perpendicular to D, and graph o with AD as diameter intersects AB, AC with E and f respectively

It is proved that ∠ EAF + ∠ EDF = 180?
∵ ad is the diameter
﹤ AED = ∠ AFD = 90 °. (the circumference angle to which the diameter corresponds is a right angle)
Ψ AED + ∠ AFD = 180 °, EAF + ∠ EDF = 360 ° - (∠ AED + ∠ AFD) = 180 ° (sum of inner angles of quadrilateral is 360 degrees)

As shown in the figure, ad, a'd 'are the heights on the sides BC and b'c' of the acute triangle ABC and a'b'c ', respectively, and ab = a'B', ad = a'd '. Please add a condition that △ ABC is all equal to △ a'b'c', and explain the reasons

CD = C "d", RT triangle abd is equal to RT triangle a "B" d ". Obviously, both sides are equal, or right angle. If CD = C" d ", then the ADC of RT triangle is equal to RT triangle a" d "C". Then two triangles share one side, which is of course congruent. In fact, this is the common sense of thinking. It can also be proved by normal methods that the three sides are obviously equal

If the triangle ABC is inscribed in the circle 0, the angle ABC = 120 degrees, ab = AC, BD is the diameter of circle 0, ad = 6, then BC =? If the triangle ABC is inscribed in circle 0, the angle BAC = 120 degrees, ab = AC, BD is the diameter of circle 0, ad = 6, then BC =?

If AB = AC, then the angle ABC = angle ACB, but the angle ABC = 120 degrees, it is impossible to mark the wrong angle... Add: connect Ao, intersect BC with E, because AB = AC, we can prove that Ao is the bisector of angle BAC, and it is vertical and bisector BC. Because BD is the diameter, so the trigonometric bad is a right triangle, and can be

In the inscribed triangle ABC, BC is the diameter, AC = 4, ab = 3, ad bisection angle BAC intersects circle D, connect BD, find the length of BD?

According to Pythagorean theorem, BC = 5
∵ ad bisection angle BAC intersects with D
/ / arc BD = arc CD
∴BD=CD
∵ BC is the diameter
∴∠BAC=90°
The △ BCD is an isosceles right triangle
/ / BD = (5 / 2) radical 2

As shown in the figure, in the triangle ABC, ad bisects the angle BAC, be perpendicular to point E and intersects ad at point F. try to explain that angle 2 = 1 / 2 (angle ABC + angle c)

prove:
Because in triangle ABC,
Angle ABC + angle c = 180 degrees - angle a
Because of ad bisection angle A
So 1 / 2 (angle ABC + angle c) = 1 / 2 (180 degrees - angle a) = 90 degrees-
Because be vertical AC
So the angle BEA is 90 degrees
So angle 2 = 90 degrees - angle DAC
So angle 2 = 1 / 2 (angle ABC + angle c)

As shown in Fig. 1, in △ ABC, ab = AC, point D is the midpoint of BC, and point E is on AD (1) Results: be = CE; (2) As shown in Fig. 2, if the extension of be intersects AC at point F, and BF ⊥ AC, the foot perpendicular is f, ∠ BAC = 45 ° and other conditions remain unchanged

It is proved that: (1) ∵ AB = AC, D is the midpoint of BC,
∴∠BAE=∠EAC，
In △ Abe and △ ace,
AB＝AC
∠BAE＝∠EAC
AE＝AE ，
∴△ABE≌△ACE（SAS），
∴BE=CE；
（2）∵∠BAC=45°，BF⊥AF，
The △ ABF is an isosceles right triangle,
∴AF=BF，
∵ AB = AC, point D is the midpoint of BC,
∴AD⊥BC，
∴∠EAF+∠C=90°，
∵BF⊥AC，
∴∠CBF+∠C=90°，
∴∠EAF=∠CBF，
In △ AEF and △ BCF,
∠EAF＝∠CBF
AF＝BF
∠AFE＝∠BFC＝90° ，
∴△AEF≌△BCF（ASA）．

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

It is proved that: ∵ ad ⊥ BC,  BDA = 90 °, ∵∵ BAC = 90 °,  ABC + ∠ C = 90 °, ABC + ∠ bad = 90 °, ∵ bad = ∠ C, ? an bisection ? DAC, ? can = ∠ Dan, ? ban = ∠ bad + ∠ Dan, ? be ⊥ an

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

Proof: ∵ ad ⊥ BC,
∴∠BDA=90°，
∵∠BAC=90°，
∴∠ABC+∠C=90°，∠ABC+∠BAD=90°，
∴∠BAD=∠C，
∵ an bisection ∠ DAC,
∴∠CAN=∠DAN，
∵∠BAN=∠BAD+∠DAN，∠BNA=∠C+∠CAN，
∴∠BAN=∠BNA，
∵ be bisection ∵ ABC,
∴BE⊥AN，OA=ON，
Similarly, OM = OE,
The quadrilateral amne is a parallelogram,
The parallelogram amne is a diamond

As shown in the figure, in triangle ABC, ad bisector angle BAC, its extension line intersects the circumscribed circle O of triangle ABC at point h, and then makes EF parallel BC intersection through H AC.AB The extension line of E.F If ah = 8, DH = 2, find ch =?

You look at the picture, because ad bisects the angle BAC, and it is circumscribed, so ∠ bad and ∠ BCH are on the same arc. So ∠ bad = ∠ CAD = ∠ BCH
So it is easy to prove that △ AHC ∽ CHD, so ch ∽ DH × ah = 2 × 8 = 16. So ch = 4
EF ‖ BC, this condition is useless

As shown in the figure, ad is the diameter of the circumscribed circle of the triangle ABC, ad is perpendicular to BC, the perpendicular foot is point F, and the bisector of ∠ ABC intersects ad at point E, connecting BD and CD As shown in the figure, ad is the diameter of the circumscribed circle of the triangle ABC. Ad is perpendicular to BC, and the perpendicular foot is point F. the bisector of ∠ ABC intersects ad at point E and connects BD and CD (1) verification: BD = CD; (2) please judge whether points B, e and C are on the circle with D as the center and DB as the radius? Please explain the reasons

Analysis: (1) it can be proved by equal arc and equal chord
(2) In this paper, we use the equal arc to make the circumference angles equal, and then we can get ∠ DBE = ∠ DEB by the equivalent substitution of ∠ bad = ∠ CBD, thus proving that DB = de = DC, so B, e, C are on the circle with D as the center and DB as the radius
It is proved that: (1) ∵ ad is the diameter, ad ⊥ BC,
∴ BD^=CD^
∴BD=CD．
(2) B, e and C are on the circle with D as the center and DB as the radius
Reason: it is known from (1) that BD ^ = CD ^,
∴∠BAD=∠CBD,
And ∵ be bisection ∠ ABC,  CBE = ∠ Abe,
∵∠DBE=∠CBD+∠CBE,∠DEB=∠BAD+∠ABE,∠CBE=∠ABE,
∴∠DBE=∠DEB,
∴DB=DE．
According to (1), BD = CD
∴DB=DE=DC．
The three points B, e and C are on the circle with D as the center and DB as the radius