As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC at D, ﹤ bad = 40 °, point E on AC, and AE = ad, calculate the degree of ∠ CDE ditto

As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC at D, ﹤ bad = 40 °, point E on AC, and AE = ad, calculate the degree of ∠ CDE ditto

∵ AB = AC, ad ⊥ BC in D, ad = ad
≌ delta ADC
∴CAD=40
∵AE=AD
∵∠ADE=∠AED=70
∠CDE=20

In the triangle ABC, ab = AC, D, e are points on BC and AC respectively. What condition does the angle bad and angle CDE satisfy? Ad = AE

Let ∠ CDE = α, ∠ bad = β,
If ad = AE, there should be ∠ AED = ∠ ade,
From ∠ AED = ∠ C + α (1)
∠ADE+α=β+∠B（2）
∵∠B=∠C,
The results of (2) - (1) are as follows:
α=β-α,
∴β=2α,
In other words, bad = 2 CDE

As shown in the figure, triangle ABC and triangle CDE are equilateral triangles. Points E and F are on AC and BC respectively, and AE and BF intersect at point G 1. Find the degree of angle AGB 2. Link DG and verify that DG = Ag + BG

If be = CF, the conclusion can hold. 1, be = CF ∠ Abe = ∠ BCF = 60 AB = BC △ Abe ≌ △ BCF ∠ BAE = ∠ CBF ∠ AG

As shown in the figure, in ▱ ABCD, ∠ DAB = 60 ° and points E and F are respectively on the extension line of CD and AB, and AE = ad, CF = CB (1) It is proved that the quadrilateral afce is a parallelogram; (2) If "DAB = 60 °" is removed from the known condition, is the above conclusion still valid? If yes, please write the proof process; if not, please explain the reasons

(1) It is proved that:

As shown in the figure, ab = AC, point D is the midpoint of BC, AB bisector angle DAE, AE is vertical be, and perpendicular foot is e

What's the problem

As shown in the figure, in the square ABCD, point E is a point on BC, AF bisection angle DAE intersects CD at F, and AE = be + DF is proved

Extend CD to h so that DH = be,
By be + FD = FH, AE = ah, as long as ah = FH
By △ Abe ≌ △ ADH, (SAS)
∴AE=AH（1）
From ∠ BAF = ∠ HAF,
Ab ∥ CD,  ABF = ∠ AFH,
It is concluded that: ∠ HAF = ∠ AFH,
∴HF=AH=AE,
That is, AE = be + DF is correct

It is known that: as shown in the figure, ab = AC, point D is the midpoint of BC, AB bisection ∠ DAE, AE ⊥ be, and the perpendicular foot is e Confirmation: ad = AE

It is proved that: ∵ AB = AC, point D is the midpoint of BC,
∴∠ADB=90°，
∵AE⊥EB，
∴∠E=∠ADB=90°，
∵ AB bisection ∠ DAE,
∴∠1=∠2；
In △ ADB and △ AEB,
∠E＝∠ADB＝90°
∠1＝∠2
AB＝AB ，
∴△ADB≌△AEB（AAS），
∴AD=AE．

As shown in the figure, M is the midpoint of be, ab = AC, ad = AE, and am is perpendicular to CD The common vertex of triangle Abe and triangle ADC is a,

The intersection of AM and CD is the point n, extend am to F, so that MF = am ? BM = EM ᙽ the abafe is a parallelparallelparallelparallelogram ᙨ BF = AE  Abf  BAE = 180 ∵ BAC = ? DAC = ∵ DAE = 90 ∵ CAD ? ad = 90 ? CAD ? BF = AE ad = AE ? BF = ad = ad ? AB = AB it's a good idea

As shown in the figure, ab = AC, ad = AE, M is the midpoint of be, angle BAC = angle, DAE = 90 °. Verification: am vertical DC

0

The triangle ABC, ab = AC, ad is perpendicular to D, f is the midpoint of AD and intersects AC at E. how much AC is AE equal to?

DP ∥ be is made by D, and DP and AC are delivered to P
∵ AB = AC, ad is perpendicular to d
∴BD=CD
∵DP‖BE
∴PC/EP=CD/DB=1
∴PC=EP
∵ f is the mid point of AD, DP ∥ be
∴AE=EP
∴AE=EP=PC
∴AE=AC/3