As shown in the figure, in △ ABC, it is known that ab = BC = Ca, AE = CD, ad and be intersect at point P, BQ ⊥ ad at point Q. verification: BP = 2pq

As shown in the figure, in △ ABC, it is known that ab = BC = Ca, AE = CD, ad and be intersect at point P, BQ ⊥ ad at point Q. verification: BP = 2pq

Proof: ab = BC = ca,
ν Δ ABC is an equilateral triangle,
∴∠BAC=∠C=60°，
In △ Abe and △ CAD
AB＝AC
∠BAC＝∠C
AE＝DC
∴△ABE≌△CAD（SAS），
∴∠ABE=∠CAD，
∵∠BPQ=∠ABE+∠BAP，
∴∠BPQ=∠CAD+∠BAP=∠CAB=60°，
∵BQ⊥AD
∴∠BQP=90°，
∴∠PBQ=30°，
∴BP=2PQ．

As shown in the figure, in △ ABC, it is known that ab = BC = Ca, AE = CD, ad and be intersect at point P, BQ ⊥ ad at point Q. verification: BP = 2pq

It is proved that:

In △ ABC, ad is the center line, O is the middle point of AD. the straight line a passes through point O and passes through points a, B and C to make the vertical lines of line a respectively, and the vertical feet are g, e and F. when the line a rotates around point O to be perpendicular to ad (as shown in Figure 1), it is easy to prove that be + CF = 2ag, When the line a rotates around point O and is not perpendicular to ad, what is the quantitative relationship between the segments be, CF and Ag in the two cases of Fig. 2 and Fig. 3? Please write your conjecture and prove it in Figure 3

(1) Conjecture results: the conclusion of Figure 2 is be + CF = 2ag, and the conclusion of Figure 3 is be-cf = 2ag. (2) it is proved that connecting CE, passing D is DQ ⊥ L, perpendicular foot is Q, intersection CE is h (Fig. 4), ? ago = ∠ dqo = 90 °, AOG = ∠ DOQ (equal to vertex angle), and O is the midpoint of AD, namely Ao = do, ≌≌ △ DOQ (AAS

As shown in Fig Verification ∠ CAF = ∠ B

prove:
∵ ad bisection ∵ bac
∴∠BAD=∠CAD
∵DE‖AC
∴∠EDA=∠CAD
∴∠EAD=∠EDA
∴EA=ED
∵EF⊥AD
ν EF vertically bisects ad
∴FA=FD
∴∠FAD=∠FDA
∴∠FAC+∠CAD=∠B+∠BAD
∵∠BDA=∠CAD
∴∠CAF=∠B

As shown in the figure, △ ABC is an isosceles right triangle, ∠ ACB = 90 ° and ad is the center line on the side of BC. Through C, the vertical line of ad is made, which intersects AB at point E and ad at point F. it is proved that ∠ ADC = ∠ BDE

Make ch ⊥ AB at h and ad at P, ∵ in RT ⊥ ABC, AC = CB, ∵ ACB = 90 °, cab = ∠ CBA = 45 °. ∵ HCB = 90 ° - ∵ CBA = 45 ° = ∵ BC midpoint is D, ? CD = BD. and ? ch ⊥ AB, ∵ ch = ah = BH. And ? PAH + ∠ APH = 90 °, PCF + ∠ CPF = 90 °, APH =

As shown in the figure, the area of the triangle ABC is 54 square centimeters, be: EC = 1:2, ad: DB = 1:2. Find the area of the triangle ade. (AE is the BC vertical line.) SOS

Triangle Abe is 1 / 3 of the area of triangle ABC (because BC = 3bE)
Triangle ade is 1 / 3 of triangle Abe area (because AB = 3aD)
So the triangle ade is (1 / 3) * (1 / 3) = 1 / 9 of triangle ABC
54*(1/9)=6
So it's six centimeters square
Primary school problems

As shown in the figure, be ⊥ ad, CF ⊥ AD are known, and be = cf. please judge whether AD is the center line of ⊥ ABC or the angular bisector? Please state the reasons for your judgment

Ad is the midline of △ ABC
The reasons are as follows.
∵BE⊥AD，CF⊥AD，
∴∠BED=∠CFD=90°，
In △ BDE and △ CDF,
∠BED＝ ∠CFD
∠BDE＝∠CDF
BE＝CF
∴△BDE≌△CDF（AAS），
∴BD=CD．
The ad is the midline of △ ABC

As shown in the figure, in △ ABC, points D, e and F are on three sides respectively, e is the midpoint of AC, ad, be and CF intersect at a point G, BD = 2dc, s △ GEC = 3, s △ GDC = 4, then the area of △ ABC is () A. 25 B. .30 C. 35 D. 40

BD=2DC，
∴S△ABD=2S△ACD，
∴S△ABC=3S△ACD，
∵ e is the midpoint of AC,
∴S△AGE=S△CGE，
And ∵ s △ GEC = 3, s △ GDC = 4,
∴S△ACD=S△AGE+S△CGE+S△CGD=3+3+4=10，
∴S△ABC=3S△ACD=3×10=30．
Therefore, B

As shown in the figure, in △ ABC, O is the intersection point of high AD and be. Observe the graph and try to guess what kind of quantitative relationship exists between ∠ C and ∠ doe, and prove your conjecture conclusion

∠C+∠DOE=180°．
∵ ad, be is the height of ∵ ABC (known),
Ψ AEO = ∠ ADC = 90 ° (high meaning),
∵ DOE is the external angle of  AOE (the concept of triangle external angle),
Ψ DOE = ∠ OAE + ∠ AEO (one outer angle of a triangle is equal to the sum of two nonadjacent interior angles)
=∠OAE+90°（∠AEO=90°）
=∠OAE+∠ADC（∠ADC=90°）
∴∠C+∠DOE=∠OAE+∠C+∠ADC=90°+90°=180°．
Another method: in the quadrilateral ceod, ∠ C + ∠ EOD + 90 ° + 90 ° = 360 °,
Then ∠ C + ∠ EOD = 180 °

As shown in the figure, it is known that in △ ABC, ad is high, CE is midline, DC = be, DG ⊥ CE, G is perpendicular foot It is proved that: (1) g is the midpoint of CE; (2) B = 2 ∠ BCE

It is proved that: (1) connect De;
∵ ad ⊥ BC, e is the midpoint of ab,
﹤ De is the center line on the oblique side of RT △ abd, i.e., de = be = 1
2AB；
∴DC=DE=BE；
And ∵ DG = DG,
∴Rt△EDG≌Rt△CDG；（HL）
∴GE=CG，
/ / G is the midpoint of CE
(2) According to (1), be = de = CD;
∴∠B=∠BDE，∠DEC=∠DCE；
∴∠B=∠BDE=2∠BCE．