Senior two math proof questions 1, as shown in the figure, AB=AC,? BAC=90？ ，BD？ AE, CE in d AE in e. and BD> church of England
It is proved that BD=EC+ED.
Answer: Proof: ∵? BAC=90？ ，CE？ AE，BD？ AE，
ABD+？ Bad =90? ,? Bad+? DAC=90？ ,? ADB=？ AEC=90？ .
ABD=？ DAC。
AB = AC，(
? △ABD?△CAE(AAS)。
? BD=AE，EC=AD。
∫AE = AD+DE，
? BD=EC+ED。
△ ABC is an equilateral triangle. ? ACB=90？ , AD is the center line on the side of BC, the vertical line passing through C is AD, the point passing through AB is E, and the point passing through AD is F. Verify? ADC=？ BDE
Solution: as CH? AB crosses AD at h and p,
∫AC = CB in Rt△ABC，？ ACB=90？ ,
CAB=？ CBA=45？ .
HCB=90？ -? CBA=45？ =? CBA。
And: the midpoint d,
? CD=BD。
CH again? AB，
? CH=AH=BH。
Again? PAH+？ APH=90？ ,? PCF+？ CPF=90？ ,? APH=？ Central provident fund,
PAH=？ PCF。
Again? APH=？ CEH，
At △APH and △CEH,
? PAH=？ ECH, ah =CH,? PHA=？ EHC，
? △APH?△CEH(ASA)。
? PH=EH，
∫PC = CH-PH，BE=BH-HE，
? CP=EB。
In delta △PDC and delta △EDB
PC=EB，？ PCD=？ DC=DB，
? △PDC?△EDB(SAS)。
ADC=？ BDE。
2
Proof: OE? AB, OF in e? AC in f,
∵? 3=? 4,
? OE=OF。 (Here comes the question. What is the reason? I'm a little confused. )
∵? 1=? 2,
? OB=OC。
? Rt△OBE≌Rt△OCF(HL)。
5=? 6.
1+? 5=? 2+? 6.
Namely. ABC=？ ACB。
? AB=AC。
? △ABC is an isosceles triangle
A little o stands for OD AB Yu d
Give OE some o? AC Yu e
Re-proving Rt△AOD≌ Rt△AOE(AAS)
Get OD=OE.
You can prove Rt△DOB≌ Rt△EOC(HL) again.
Understand? ABO=？ AdaptiveControlOptimization (adaptive control optimization)
Because again? OBC=？ Oil circuit breaker (oil circuit breaker)
Understand? ABC=？ American Broadcasting Company Inc (ABC)
Get isosceles triangle △ABC
4. 1.E is a point of ray AB, square ABCD and square DEFG have a common vertex d, when e is moving,? Is the size of FBH a fixed value? And verify
(more than f is FM? AH is in m, and △ADE is equal to △MEF.
2. Triangle ABC, with AB and AC as sides, makes square ABMN and square ACPQ.
1) If DE? In BC, it was proved that e was the midpoint of NQ.
2) If d is the midpoint of BC,? BAC=90？ , verification: AE? NQ
3) if f is the midpoint of MP, FG? BC in G, verification: 2FG=BC.
3. it is known that AD is the height on the edge of BC, and BE is? The bisector of ABC, EF? B.C. in F, and after A.D. in G.
Proof: 1)AE=AG (this proof is good) 2) The quadrilateral AEFG is a diamond.
4. in quadrilateral ABCD, AB=DC,? B=？ C & lt90. Verification: The quadrilateral ABCD is trapezoidal.
Prove:
5. As shown in the figure: When the size is 6? In a square grid of 5, the vertices A, B and C of △A, B and C are on the vertices of the unit square. Please answer the following questions:
(1) Draw a △DEF in the diagram, so that △DEF∽△ABC (similarity ratio is not 1) requires that points D, E and F must be on the vertices of the unit square (used vertices can be used);
(2) Write the proportional expressions of their corresponding edges; And find the similarity ratio of △DEF and △ABC.
6. AB=4, AC=2, D is the midpoint of BC, and AD is an integer. Look for advertisements.
7. it is known that d is the midpoint of AB. ACB=90？ Verification: CD= 1/2AB.
8. Known: BC=DE,? B=？ e，？ C=？ D and f are the midpoint of the CD, so verify:? 1=? 2.
9. as shown in the figure: AB=AC, ME? AB，MF？ AC，e，f，ME=MF。
Verification: MB=MC
10. As shown in the figure, five equivalence relations are given: ① ad = bc2ac = bd3ce = de4? D=？ C ⑤？ DAB=？ CBA。
Please take two of them as the conditions and one of the other three as the conclusion, draw the correct conclusion (write only one situation) and prove it.
1 1. as shown in the figure: BE? AC，CF？ AB，BM=AC，CN=AB .
Verification: (1) am = an; (2)AM？ Ann.
12. As shown in the figure, AB=DC, AC=DB and BE=CE are known. Verification: AE=DE.
13. As shown in the figure, △ABC is an isosceles right triangle. ACB=90？ , AD is the middle line of BC side, crossing c is the vertical line of AD, intersecting with AB at point e and intersecting with AD at point f,