Proof: extend the intersection of BA and CE at g point
E is the midpoint of AD.
AE= 1/2AD=BC。
FE⊥GC
FE is perpendicular bisector of BC.
So △ fge △ FCE
∠G=∠FCE
∠G=∠FEA (complementary angles of equal angles are equal)
Finite element analysis =∠FCE
∠EAF=∠FEC
therefore
△AEF∽△ECF
2. In △ABC, AB=AC= 13, BC= 10, D is the midpoint of AB, and if D is DE⊥AC of point E, the length of DE is-.
Guo a Zuo
AF⊥BC in F.
BF= 1/2BC= 10/2=5
According to Pythagorean theorem
AF？ +BF？ =AB？
AF= 12
s△ABC = 1/2×BC×AF = 1/2× 10× 12 = 60
Pass b as BG⊥AC
De Berg
D is the midpoint of AB.
DE= 1/2BG
S△ABC= 1/2×AC×BG
60= 1/2× 13×BG
Blood sugar = 120/ 13
DE= 1/2BG=60/ 13
3. As shown in the figure, it is known that in △ABC, AB=AC, AD=BD=BC, then ∠ A = _ _ _ _ _ (write the conclusion directly without proof).
As shown in the figure. It is known that in △ABC, AB=AC, AD=BD=BC, then ∠A=__36 degrees _ _ (write the conclusion directly without proof).
∠A=∠ABD
∠C=∠BDC=2∠A
∠A+∠ABC+∠C =∠A+2∠A+2∠A = 180
4. As shown in the figure, in △ABC, AB=AC, D is a point above BC, ∠ Bad = 30, E is a point above AC, and AD=AE, so find the degree of ∠EDC.
As shown in the figure, in △ABC, AB=AC, D is a point above BC, ∠ bad = 30, E is a point above AC, and AD=AE. Find the degree of < ∠EDC.
Solution: according to the meaning of the problem
AD=AE
∠ Ade =∠AED
AB=AC
∠B=∠C
∠ Ade +∠EDC=∠B+30
AED+∠EDC=∠C+30
EDC+∠C+∠EDC=∠C+30
2∠EDC=30
∠EDC= 15 degrees
5. In △ABC, AD bisects ∠BAC, DE is the middle vertical line of BC, and E is the vertical foot. If D crosses, DM is perpendicular to AB in M and DN is perpendicular to the extension line from AC to AC in N, which proves that BM=CN.
Proof: AD bisects ∠BAC
DM⊥AB,DN⊥AC
So DM=DN
Connect DC DB
DE divides BC vertically.
Then DB=DC
DM=DN
Rt△DMB≌Rt△DNC
BM=CN
6. As shown in the figure, in △ABC, ∠C is a right angle, ∠ A = 30, with AB and AC as sides respectively, so that positive △ABE and positive △ACD are outside △ABC, and de and AB intersect at F. Verification: EF=FD.
Prove:
Do EG⊥AB to e
Cross from AB to g
Connect GD and AB to H, GC.
△EBA is positive △
Then g is the midpoint of AB.
GC= 1/2AB=GA
∠GCA=∠GAC=30
∠DCA=∠DAC=60
Addition of two formulas
∠DCG =∠ Dag =90
GC=GA
Gadolinium = gadolinium
△ DCG△ Dag
∠GDC=∠GDA
The bisector whose DG is ∠CDA
therefore
We can know.
DG vertically divides AC.
H is the midpoint of communication.
GH‖BC
∠EAD=60
∠BAC=30
∠EAC = 90°
∠BCA=90
BC‖EA
GH‖AE( 1)
In the same way; In a similar way
Da (2)
According to (1)(2)
therefore
Quadrilateral ADGE is a parallelogram.
GA and DE are diagonal lines.
therefore
EF=FD
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