The vertical line CE is drawn from the vertex C of rectangle ABCD to the diagonal BD, and the perpendicular foot is e. if CE is divided into two angles with ∠ C being 1:5, then ∠ ace=

The vertical line CE is drawn from the vertex C of rectangle ABCD to the diagonal BD, and the perpendicular foot is e. if CE is divided into two angles with ∠ C being 1:5, then ∠ ace=

60°
It is known from the condition that ∠ ECD is 15 °, then ∠ BDC = 75 °
Then, ADB = 15 ° and ACB = 15 °
Then ∠ ace = 90-15-15 = 60 degree
Change ∠ ECD to 75 ° and the answer remains unchanged

As shown in the figure, it is known that the quadrilateral ABCD is a rectangle, the diagonal lines AC and BD intersect at point O, CE ‖ BD, de ‖ AC, CE and de intersect at point E, which means OE ⊥ CD
And the picture I can't say O(∩_ ∩)O~

Because CE ∥ BD, de ∥ AC, the quadrilateral ODEC is a parallelogram
It is also known that the quadrilateral ABCD is a rectangle, so od = OC
So the quadrilateral ODEC is a diamond
also OE.CD It's diagonal
So OE ⊥ CD

As shown in the figure, AC and BD are the diagonals of rectangular ABCD, ah ⊥ BD in H, CG ⊥ BD in G, AE is the bisector of ∠ bad, the extension of intersection GC and E,
Verification: BD = CE

There are no pictures

As shown in the figure, e, F, m and N are the points on the four sides of the square ABCD, and AE = BF = cm = DN. Try to judge what figure efmn is and prove your conclusion

Because square ABCD
So ad = AB = BC = CD
Angle DAB = angle ABC
And because DN = AE = BF = cm
So an = EB
Therefore, the RT triangle ane is equal to the RT triangle EBF
So angle NEA = angle EFB
NE=EF
Because EFB + Feb = 90 degrees
So angle NEA + angle Feb = 90 degrees
So Nef = 90 degrees
In the same way, it can be proved that other angles are equal to 90 degrees and other sides are equal
Then efmn is a square