Some pressing axes problems in junior maths

Topic 1:

Analysis: The known two midpoint is not connected to any effect and does not have any association with the known condition AD=BC, so the midpoint m and N are destined to be used separately, see Midpoint to find the midpoint, where to find another midpoint, to construct the auxiliary median. We can try the side of the ad and BC, found that can not take advantage of AD=BC conditions, this tells us, in the existing line to find the midpoint is a dead end, quickly change the train of thought bar. We're going to create a midpoint, like the diagonal AC of ABCD, the neutral point g of AC, the GM and the GN, and in ABC, the median line, GM, is just half the BC, and in ACD, the median line GN is just half the ad.

Proof: Connect AC, make Gn∥ad AC to G, connect mg.
∵n is the midpoint of the CD, and Ng∥ad,
∴ng= 1 2 ad,g is the midpoint of the AC,
And ∵m is the midpoint of AB,
∴MG∥BC, and mg= 1 2 BC
∵ad=bc
∴ng=gm
GNM for Isosceles
∴∠gnm=∠gmn
∵GM∥BF parallel to (the nature of the median line parallel to the third side)
∴∠gmf=∠f
∵gn∥ad
∴∠gnm=∠den
∴∠den=∠f.

Summary: The difficult point is how to find the Guide line (median line) to see the midpoint to find the corresponding auxiliary line

Topic 2:

In the figure, isosceles ABC, ab=ac,∠bac=90°,be the ∠bac AC in E, over C as cd⊥be in D, connect AD, verify: (1) ∠adb=45°; (2) be=2cd.

Analysis: Since BD is both a high and a angular split line, from the three-line integration, it is easy to think if the BA, CD extension to the Q, you can construct a isosceles bcq, and then the three-line integration to know that BD is also the midline, so cq=2cd, so that the CD twice times to make, and then you have to prove cq=be , look at the image and find the triangle where the CQ and be are located, which is the proof triangle Abe≌ ACQ

Prove:

Answer: proof: (1) ∵cd⊥be,∠bac=90°
∴a, B, C, D four-point total circle
∴∠adb=∠acb
∵ isosceles ABC, ab=ac,∠bac=90°
∴∠acb=∠abc=45°
∴∠adb=45°;
(2)
Extended BA and CD to Q
∵∠caq=∠bae=∠bdc=90°
∴∠acq+∠q=90°,∠abe+∠q=90°
∴∠acq=∠abe

"In ABE and CDE, because ∠dec=∠aeb (Dianding equals) can also draw ∠acq=∠abe"

In Abe and Acq,
∠abe=∠acq
Ab=ac
∠bae=∠caq
∴ Abe≌ ACQ (definition of similar triangles)
∴BE=CQ,
∵BD equally ∠abc
∴∠qbd=∠cbd
∵∠bdc=90°
∴∠bdc=∠bdq=90°
In Qdb and CDB.
∠qbd=∠cbd
Bd=bd
∠bdq=∠bdc
∴ Qdb≌ CDB (definition of similar triangles)
∴cd=dq
∴cq=2cd
∴be=2cd

Topic 3:

As shown in the figure, on the side of the equal ABC on the edge of the BC take a little D, as ∠ade=60°,de ∠c of the outer line in E, it is proved that ADE is equilateral.
Analysis: The condition is very ideal, because the triangle Ade already has a 60° angle, if can find another 60° angle, can prove ADE is equilateral. (if learned that four points of the total circle can be drawn immediately), only in the base of 60° Shanghai Isosceles, or to find Pang, looks like Abd and ace a bit, but the conditions are not enough; Abd and Dec. Also a bit like, but the ad and de is not the corresponding edge, also not; and the triangle DCE and Ace also a bit like, but the corresponding edge also has a problem, is not to testify the edge of the triangle. This is because all seemingly congruent triangles are not congruent, the difficulty of this problem will be so great.
So how to deal with it. Do auxiliary line construction congruent triangle since you want to make ad and de equal to the corresponding side then their diagonal also should be equal, de is the ∠dce, because the CE is the angle evenly line, so ∠ace is 60°, so ∠dce=120° ad diagonally should also be 120°, So the parallel line over D to AC is Tong ∠bpd=60° ab in P,∠bac, thus creating ∠apd=120°, equilateral BPD, and AP=CD, in APD and DCE, Ap=cd,∠apd=∠dce, One more condition. Take a look at ∠pad and ∠CDE, according to the outer theorem: ∠adc=∠pad+∠b=∠pad+60°, and ∠ADC is ∠CDE and ∠ade splicing into, and because of ∠ade=60°, so get ∠cde=∠pad proof: Cross D for AC parallel line to P AB
∴ BDP for Equilateral, BD=BP
∴ap=cd
∵∠BPD is the outer of ADP
∴∠adp+∠dap=∠bpd=60°
and ∠adp+∠edc=180°-∠bdp-∠ade=60°
∴∠adp+∠dap=∠adp+∠edc=60°
∴∠dap=∠edc (outer triangle theorem)
In ADP and Dec,
∵∠dap=∠edc Ap=dc∠apd=∠dce
∴ Adp≌ DEC (congruent triangle)
∴ad=de
∵∠ade=60°
∴ Ade is equilateral.