As shown in the figure AB = ad ∠ bad = ∠ CAE AC = AE, ab = ad

As shown in the figure AB = ad ∠ bad = ∠ CAE AC = AE, ab = ad

Shocked
There is already AB = ad in the known conditions
It is also proved that ab = ad
Just write it directly
Because AB = ad
So AB = ad

As shown in the figure, it is known that the triangle ABC and the triangle ade are isosceles triangles, ab = AC, ad = AE, and angle DAB = angle EAC, then de and BC are parallel

Yes, this is correct

It is known that: as shown in Fig. 1, in △ ABC and △ ade, ab = AC, ad = AE, ∠ BAC = ∠ DAE, and points B, a, D are on a straight line, connecting be, CD, m, n are the midpoint of be and CD respectively (1) The results showed that: ① be = CD; ② △ amn is isosceles triangle; (2) On the basis of Fig. 1, rotate △ ade clockwise 180 ° around point a, and other conditions remain unchanged to obtain the figure shown in Fig. 2; (3) Under the condition of (2), please extend the ED intersection line BC at point P in Fig. 2

(1) It is proved that: (1) the

As shown in the figure, in △ ABC and △ ade, ab = AC, ad = AE, ∠ BAC = ∠ DAE, verification: △ abd ≌ △ ace

Proof: ∵ BAC = ∵ DAE (3 points)
∴∠BAC+∠CAD=∠DAE+∠CAD，
That is ∠ EAC = ∠ DAB (4 points)
In △ AEC and △ ADB
AD＝AE
∠DAB＝∠EAC
AB＝AC ，
∴△AEC≌△ADB（SAS）．… (5 points)

It is known that: as shown in Fig. 1, in △ ABC and △ ade, ab = AC, ad = AE, ∠ BAC = ∠ DAE, and points B, a, D are on a straight line, connecting be, CD, m, n are the midpoint of be and CD respectively (1) The results showed that: ① be = CD; ② △ amn is isosceles triangle; (2) On the basis of Fig. 1, rotate △ ade clockwise 180 ° around point a, and other conditions remain unchanged to obtain the figure shown in Fig. 2; (3) Under the condition of (2), please extend the ED intersection line BC at point P in Fig. 2

(1) It is proved that: (1) the

It is known as follows: in △ ABC, △ ade, ∠ BAC = ∠ DAE = 90 °, ab = AC, ad = AE, points c, D and E are on the same line, connecting BD Verification: (1) △ bad ≌ △ CAE; (2) try to guess the special position relationship between BD and CE, and prove it

(1) It is proved that: ∵≌≌≌≌≌≌≌≌≌≌≌≌≌≌≌△ CAE (SAS) ?≌≌△ CAE (SAS). ? the special position relationship between BD and CE is BD ⊥ CE

As shown in the figure, in the triangle ABC and triangle ade, ab = AC, ad = AE, connect BD, CE, and BD = CE. Verify that the angle BAC = angle DAE

Because AB=AC, AD=AE, angle BAD= angle CAE, so angle BAD+ angle DAC= angle CAE+ angle DAC, so angle BAC= angle DAE

In △ ABC, ab = AC, △ ade, ad = AE and ∠ BAC = ∠ DAE, with BD and CE intersecting at point P The relationship between ∠ BAC and ∠ BPE No one in my class can do this

∠BAC=∠DAE
∠EAC=∠DAB
AB=AC,AD=AE,
All of △ EAC is equal to △ DAB
∠ACE=∠ABD
∠ BPE = ∠ PBC + ∠ PCB = ∠ PBC + ∠ BCA + ∠ ace = ∠ PBC + ∠ abd + ∠ BCA = ∠ ABC + ∠ ACB = 180 degrees - ∠ bac

As shown in the figure, in △ ABC and △ ade, ab = AC, ad = AE, ∠ BAC = DAE = 90 °, what are the quantitative and positional relationships of segments BD and CE? Please state the reasons

The crossing point F between BD and EC was extended,
In △ ace and △ ADB,
AE＝AD
∠EAC＝∠DAB
AC＝AB ，
∴△ACE≌△ADB（SAS），
∴BD=CE，∠AEC=∠ADB，
∵∠ADB+∠ABD=90°
∴∠ABD+∠AEC=90°
∴∠BFE=90°，
∴BD⊥CE．

As shown in the figure ∠ DAB = ∠ CAE, please add a condition:______ Make △ ABC ∽ ade

∵∠DAB=∠CAE
∴∠DAE=∠BAC
The two triangles are similar when ∠ d = ∠ B or ∠ AED = ∠ C or ad: ab = AE: AC or ad · AC = ab · AE
So the answer is: ∠ d = ∠ B (the answer is not unique)