Isosceles right triangle ABC, angle c is 90 degrees, D is a point on the side of BC, through B make the vertical line of ad at point E. find the angle AEC Sorry, I haven't learned the circle yet

Isosceles right triangle ABC, angle c is 90 degrees, D is a point on the side of BC, through B make the vertical line of ad at point E. find the angle AEC Sorry, I haven't learned the circle yet

First of all, draw your own picture
Three vertices of right triangle AEB are in common circle
The three vertices of the right triangle ABC are also in the same circle
They all take AB as the diameter, so they are the same circle
So angle AEC = angle ABC = 45 degrees
The reason is that the same arc is equal to the same angle

As shown in the figure, it is known that CB is perpendicular to AB, point E is on AB, CE bisecting angle BCD, de bisecting angle ADC, angle EDC + angle DCE = 90 °, which proves that DA is perpendicular to ab

Let ∠ ade be ∠ 1 ∠ CDE be ∠ 2
The ∠ BEC is ∠ 3 ∠ ECB is ∠ 4 ∠ ECD is ∠ 5 ∠ AED is ∠ 6
∠1=∠2 ∠4=∠5 ∠4+∠3=90° ∠3+∠6=90°
∴∠4=∠6
And ∠ 2 + ∠ 5 = 90 ° (i.e. ∠ 2 + ∠ 4 = 90 °) and ∠ 4 + ∠ 3 = 90 °
∴∠3=∠2=∠1
∴∠1+∠6=∠3+∠4=90°
According to the theorem of the sum of internal angles of triangles
∠A=90°
∴DA⊥AB

As shown in the figure, the point E is on the line AB, Da ⊥ AB, CB ⊥ AB, De, CE bisect ⊥ ADC, ⊥ BCD, ad = 2, AE = 3, EC = 32 respectively. (1) find out all the similar triangles in the graph, and prove one pair of them; (2) find out the length of ab

(1) It is proved that: ∵ Da ⊥ AB, CB ⊥ AB, ∵ ad ∥ BC, then ∵ ADC + ∥ BCD = 180 ° and ∵ de and CE divide ∵ ADC, ∵ BCD equally, ∵ 2 (∵ EDC + ∥ ECD) = 180 ° respectively, then ∵ EDC + ∥ ECD = 90 ° and ∥ Dec = 90 ° in RT △ ade and RT △ EDC, ∥ ade = ∥ EDC, ∥ ade ∥ EDC; (2) in RT △ ade, ∥ ad = 2, AE = 3 is determined by Pythagorean theorem, The results show that de = 22 + 32 = 13, ∵ △ ade ∽ BEC, ∵ bend = ECDE, then be = 3213 × 2 = 62613, ∵ AB = AE + be = 3 + 61326

Known: as shown in the figure, CB ⊥ AB, CE bisection ⊥ BCD, de bisection ⊥ CDA, ⊥ 1 + ⊥ 2 = 90 ° verification: Da ⊥ ab

∵ CE bisection ∠ BCD, de bisection ∠ CDA, ∵ 1 = 12 ‰ ADC, ∵ 2 = 12 ‰ BCD, ∵ 1 + ∵ 2 = 12 ‰ ADC + 12 ‰ BCD = 12 (∵ ADC + ∵ BCD) = 90 °, ∵ ADC + ∵ BCD = 180 °, ∵ ad ‖ BC, ∵ a + ∵ B = 180 °,