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As shown in the figure, △ ABC, ab = AC, ⊙ o with ab as diameter intersects BC at point P, PD ⊥ AC at point D (1) It is proved that PD is tangent of ⊙ o; (2) If ∠ cab = 120 ° and ab = 2, calculate the value of BC
(1) Proof: connect AP, Op,
∵AB=AC，
∴∠C=∠B，
And ∵ OP = ob, ∵ OPB = ∠ B,
∴∠C=∠OPB，
∴OP∥AD；
And ∵ PD ⊥ AC to d,
∴∠ADP=90°，
∴∠DPO=90°，
⊙ o with ab as diameter intersects BC at point P,
⊙ PD is the tangent of ⊙ o
(2) ∵ AB is the diameter,
∴∠APB=90°；
∵AB=AC=2，∠CAB=120°，
∴∠BAP=60°，
∴BP=
3，
∴BC=2
3．
As shown in the figure, in △ ABC, ⊙ o with ab diameter intersects BC at point P, PD ⊥ AC at point D, and PD is tangent to ⊙ o (1) Results: ab = AC; (2) If BC = 6, ab = 4, find the value of CD
(1) It is proved that: connect OP, ∵ PD and ⊙ o tangent, ? op ⊥ PD, ∵ AC ⊥ PD, ∵ op ∵ AC, ∵ OP = 0A = ob = 12ab, ? OP is the median line of △ ABC, ? OP = 12ac, ? AC = ab. (2) connect AP, ? AB is the diameter, ? AP X BC; from (1), AC = AB = 4, ? PC = Pb; and ? BC = 6
As shown in the figure, ad and ad 'are the heights on the edges BC and B ′ C ′ of the acute angles △ ABC and △ a ′ B ′ C ′ respectively, and ab = a ′ B ′ and ad = a ′ D ′. If △ ABC ≌ △ a ′ B ′ C ′, please add some conditions______ (fill in only one) and prove it
Additive condition: DC = d'c '
It is proved that ∵ AB = a ′ B ′, ad = a ′ D ′, ∵ ADB = ∠ a ′ D ′ B ′ = 90 °,
∴Rt△ADB≌Rt△A′D′B′（HL），
∴∠B=∠B′，BD=B′D′，
∵DC=D′C′，
∴BC=B′C′；
And ∵ AB = a ′ B ′,
∴△ABC≌△A′B′C′（SAS）．
It can also be added as follows: AC = a ′ C ′, or ∠ C = ∠ C ′, or AC = a ′ C ′, or ∠ DAC = ∠ D ′ a ′ C ′
In the acute triangle ABC, height ad = 12, edge AC = 13, BC = 14, find the length of ab
As shown in the figure:
∵ height ad = 12, edge AC = 13,
According to Pythagorean theorem, CD=
AC2−AD2=
132−122=5，
∵BC=14，
∴BD=14-5=9，
In RT △ abd, ab=
AD2+BD2=
122+92=15．
As shown in the figure, in the acute triangle ABC, ad ⊥ BC, ad = 12, AC = 13, BC = 14=______ .
∵AD⊥BC，
In RT △ ACD, CD=
AC2−AD2=
132−122=5，
∵BC=14，∴BD=BC-CD=9，
In RT △ abd, ab=
BD2+AD2=
92+122=15．
So the answer is: 15
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 ' A.BC=B'C' B.AC=A'C' C.∠C=∠C' D.∠BAC=∠B'A'C' Single choice
No, B (SSA)
The reason for a is (SAS)
The reason for C is (AAS)
The reason for D is (ASA)
As shown in the figure, there is a piece of triangle material with acute angle, with side BC = 120mm and height ad = 80mm. If it is to be processed into a square part, with one side on BC and the other two vertices on AB and AC, the side length of the square part is () A. 40mm B. 45mm C. 48mm D. 60mm
Let the side length of the square be XMM,
Then AK = ad-x = 80-x,
∵ efgh is square,
∴EH∥FG，
∴△AEH∽△ABC，
∴EH
BC=AK
AD，
That is X
120=80-x
80，
X = 48mm,
Therefore, C
If the triangle ABC is inscribed in the circle O, ab = BC, angle ABC = 120, ad = 0, ad = 6, then BD =?
∵AB=BC,∠ABC=120°,
∴∠ACB=30°．
∴∠ADB=∠ACB=30°．
∵ ad is the diameter of ⊙ o,
∴∠ABD=90°,
∴BD=AD•cos30°=3√3
As shown in the figure, in the triangle ABC, the angle Abe = 2, angle c, ad is the bisector of angle BAC, be is perpendicular to ad, and the perpendicular foot is e; If the angle c is not equal to 30 degrees, be = 1 / 2 (AC AB)
(1) Proof: because be is perpendicular to E
So the angle BEA is 90 degrees
Because the angle c is 30 degrees
Angle Abe = 2 angle c
So the angle Abe is 60 degrees
Because angle Abe + angle bea + angle BAE = 180 degrees
So the angle Bae is 30 degrees
So in a right triangle Abe, the angle BEA is 90 degrees and the angle Bae is 30 degrees
So be = 1 / 2Ab
So AB = 2be
(2) It is proved that extended be intersects AC at F
Because ad is the bisector of the angle BAC
So the angle FAE = the angle BAE
Because be is perpendicular to E
So the angle AEF = angle AEB = 90 degrees
Because AE = AE
So triangle AEF and triangle AEB are congruent (ASA)
So AF = ab
EF=BE=1/2BF
Angle AFE = angle Abe
Because angle AFE = angle c + angle CBF
Angle Abe = 2 angle c
So angle c = angle CBF
So CF = BF = 2be
Because AC = AF + CF
So AB + 2be = AC
So be = 1 / 2 (ac-ab)
As shown in the figure, in △ ABC, ad bisects ∠ BAC, ad = AB, cm ⊥ ad intersects ad extension line at point M. verification: am = 1 2（AB+AC）．
It is proved that: extend am to N, make DM = Mn, connect CN,
∵CM⊥AD，DM=MN，
∴CN=CD，
∴∠CDN=∠DNC，
∴∠DNC=∠ADB，
∵AD=AB，
∴∠B=∠ADB，
∴∠B=∠ANC，
∵∠BAD=∠CAD，
∴∠ADB=∠ACN，
∴∠ANC=∠ACN，
∴AN=AC，
∴AB+AC=AD+AN=AD+AM+MN=AD+AM+DM=2AM，
∴AM=1
2（AB+AC）．
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