As shown in the figure, point E is a point in the square ABCD, connecting AE, be and CE. Rotate △ Abe clockwise 90 ° around point B to the position of △ CBE '. If AE = 1, be = 2 and CE = 3, then ∠ be ′ C =... Degree

As shown in the figure, point E is a point in the square ABCD, connecting AE, be and CE. Rotate △ Abe clockwise 90 ° around point B to the position of △ CBE '. If AE = 1, be = 2 and CE = 3, then ∠ be ′ C =... Degree

∵ rotation,
∴BE=BE'=2,∠EBE'=90°,
∴∠BE'E=45°,EE'=2√2,
E'c = EA = 1, CE = 3,
∴E'E²+E'C²=CE²,
∴∠EE'C=90°,
∴∠BE'E=135°
I hope it can help you

E is a point in the square ABCD, and the triangle EAB is an equilateral triangle
In RT triangle ABC, ∠ ACB = 90 ° de bisects BC vertically and D is the perpendicular foot. In RT triangle ABC, ∠ ACB = 90 ° de bisects BC vertically and D is the perpendicular foot,
It is proved that ACEF is a parallelogram

I'm curious. What do you want to prove?
If D is perpendicular, D is on the BC side of the RT triangle
How could it be square ABCD
Don't you understand?
Make it clear

As shown in the figure, for square ABCD, there is a point E, AE = 3, EB = 1 on the edge of AB, and there is a point P on AC, so that EP + BP is the shortest, then the shortest distance of EP + BP is______ ．

Connect De, intersect line AC at point P, ∵ quadrilateral ABCD is square, ∵ B, D are symmetrical about line AC, ∵ De's length is the shortest distance of EP + BP, ∵ de = ad2 + AE2 = 42 + 32 = 5

In square ABCD, e is a point on diagonal AC, ab = AE, P is any point on EB. PF is perpendicular to AB
In square ABCD, e is a point on the diagonal AC, ab = AE, P is any point on EB. PF is perpendicular to AB, PG is perpendicular to AC, and the perpendicular feet are f and g respectively
Come on!

We can't draw a picture. Let's just say that the line EM is perpendicular to AB and the perpendicular foot is m. on AB, the line BN is perpendicular to AC and the perpendicular foot is n. on AC, according to the similar triangle, we can get: PF / EM = Pb / EB and pg / BN = PE / EB. Because the angle cab, ABN and AEM are all 45 degrees, we can get: EM = aesin45, BN = absin45, because AE = AB