It is known that, as shown in the figure, in △ ABC, ad and AE are the bisectors of the height and angle of △ ABC respectively, if ∠ B = 30 ° and ∠ C = 50 ° (1) Find the degree of ∠ DAE; (2) What is the relationship between ∠ DAE and ∠ C - ∠ B? (no need to prove)

It is known that, as shown in the figure, in △ ABC, ad and AE are the bisectors of the height and angle of △ ABC respectively, if ∠ B = 30 ° and ∠ C = 50 ° (1) Find the degree of ∠ DAE; (2) What is the relationship between ∠ DAE and ∠ C - ∠ B? (no need to prove)

（1）∵∠B=30°，∠C=50°，
∴∠BAC=180°-30°-50°=100°．
∵ AE is the bisector of ∵ BAC,
∴∠BAE=50°．
In RT △ abd, ∠ bad = 90 ° - ∠ B = 60 °,
∴∠DAE=∠BAD-∠BAE=60°-50=10°；
（2）∠C-∠B=2∠DAE．

It is known that AD and AE are the midline and bisector of the triangle ABC respectively A. The distance from point B and point C to AE is equal B. Point e to AB.AC The distance is equal C. The distance from B and C to D is equal D. The distance from B and C to ad is equal

A is wrong and cannot be proved
C is right without thinking about it
According to the title, draw a picture and use two triangles congruence (AAS) to know that B (proved by the AAS of right angles, bisectors and common sides) and D (AAS proof of the equality of right angles, opposite vertex angles and midpoint bisectors) are correct;

As shown in the figure, in the triangle ABC, AC = BC, AC is perpendicular to BC, D is the midpoint of BC, CF is perpendicular to ad, BF is parallel to AC, AB is vertically bisected DF

∵AC＝BC,
∴CAB＝CBA
∵BF‖AC
∴∠CAB＝ABF
∴CBA＝ABF
∵ a = 45 degrees
/ / CBA = ABF = 45 degrees
ν CBA + ABF = 90 degrees = ACB
And ∵ CDE ∵ ADC
∴CAD＝FCB
∴ACD≌CBF
∴CD＝BF＝BD
It is proved that BF = BD, CBA = ABF, and there is a common side. We know that BDX ≌ bfx (x is the intersection point of Ba and DF)
ν AB vertical bisection DF

As shown in the figure, in △ ABC, ad ⊥ BC, be ⊥ AC, the vertical feet are D and e respectively, ad and be intersect at point F, if BF = AC, calculate the size of ⊥ ABC

∵ ad ⊥ BC, be ⊥ AC (known), ⊥ ADC = ⊥ BEC = 90 ° (vertical definition), and ∵∵ AFE = ∠ BFD (equal to vertex angle), ∵ AEF ∵ BDF (two pairs of triangles with equal corresponding angles are similar), ∵ FAE = ∠ FBD (corresponding angles of similar triangles are equal), in △ BFD and △ ACD

As shown in the figure, ad is the height of triangle ABC, e is a point on AC, be intersects ad with F, and BF = AC, FD = CD, be is be perpendicular to AC You can draw your own pictures

vertical
BF = AC, FD = CD and ad is perpendicular to BC, so triangle BFD and triangle ACD are congruent
So the angle BFD equals the angle c equals the angle AFE
Angle C plus angle CAD equals 90 degrees
Then angle AFE plus angle CAD is equal to 90 degrees
Be is perpendicular to AC

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:

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:

0

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．

The geometric proof is shown in the figure. In the equilateral triangle ABC, points D and E are on the sides BC and Ca respectively, so that CD = AE, ad and be intersect at point P, BQ is perpendicular to AD and point Q, Find QP / QB value

∵ △ ABC is an equilateral triangle
∴∠C=∠BAC=∠BAE=60°
AB=AC
∵AE=CD
∴△ABE≌△ACD
∴∠CAD=∠ABE=∠ABP
∵∠BAD+∠CAD=60°
That is, BAP + CAD = 60 degrees
∴∠BAP+∠ABP=60°
∴∠BPD=∠BPQ=∠ABP+∠BAP=60°
∵BQ⊥AD
/ / in RT △ bpq
∠PBQ=90°-∠BPQ=30°
tan∠PBQ=QP/QB
∴QP/QB=tan30°=√3/3

As shown in the figure, △ ABC and △ ECD are equilateral triangles on a straight line at points B, C and D. It is proved that be = ad

It is proved that ∵ ABC and △ ECD are equilateral triangles,
∴∠ACB=∠ECD=60°，BC=AC，EC=CD．
∴∠ACB+∠ACE=∠ECD+∠ACE，
That is ∠ BCE = ∠ ACD
In △ BCE and △ ACD,
BC＝AC
∠BCE＝∠ACD
EC＝CD
∴△BCE≌△ACD（SAS）．
ν be = ad. (the corresponding sides of an congruent triangle are equal)