As shown in the figure, in rectangular ABCD, diagonal lines AC and BD intersect at point O, AE ⊥ at BD and E. if ∠ DAE = 3 ∠ BAE, try to find the degree of ∠ EAC

As shown in the figure, in rectangular ABCD, diagonal lines AC and BD intersect at point O, AE ⊥ at BD and E. if ∠ DAE = 3 ∠ BAE, try to find the degree of ∠ EAC

∵∠DAE=3∠BAE,∠DAE+∠BAE=∠BAD=90°
∴∠BAE=90°/4=22.5°
∵AE⊥BD
∴∠BAC=∠ABD=90°-∠BAE=67.5°
∴∠EAC=∠BAC-∠BAE=45°

In rectangular ABCD, AE ⊥ BD is at point E, ∠ DAE = 3 ∠ BAE?

45 degrees

As shown in the figure, in the parallelogram ABCD, 2Ab = ad, ab = AE = BF, verification: EC ⊥ FD

Let AB = x, let EC intersect ad with m FD intersect BC with N, let 2Ab = ad get ad = 2x, let AB = AE = BF get AE = BF = x, then AF = AB + BF = 2x, that is AF = ad triangle, AFD is isosceles triangle, BN is median line of the triangle, be = BC = 2x triangle, BCE is isosceles triangle, and am is median line of the triangle, then am = x, so triangle

In the parallelogram ABCD, ad = 2Ab, extend AB to two sides, make AE = BF = AB, which means that EC is vertical

Let CE intersect ad with G and DF intersect BC with H
It is easy to prove that Ag = BH = 0.5ad = ab,
So the parallelogram ghcd is rhombic, of course the diagonal is vertical