As shown in the figure, it is known that in △ ABC, ab = AC, ad is the square line of ∠ BAC, AE is the square line of the outer corner of ∠ BAC, CE is perpendicular to point E, wait online for 5 minutes, high score As shown in the figure, it is known that in △ ABC, ab = AC, ad is the square line of ∠ BAC, AE is the square line of the outer corner of ∠ BAC, and CE is perpendicular to point E (1) The quadrilateral adce is a rectangle (2) Quadrilateral ABDE is parallelogram

As shown in the figure, it is known that in △ ABC, ab = AC, ad is the square line of ∠ BAC, AE is the square line of the outer corner of ∠ BAC, CE is perpendicular to point E, wait online for 5 minutes, high score As shown in the figure, it is known that in △ ABC, ab = AC, ad is the square line of ∠ BAC, AE is the square line of the outer corner of ∠ BAC, and CE is perpendicular to point E (1) The quadrilateral adce is a rectangle (2) Quadrilateral ABDE is parallelogram

(1) Because ad is the square line of ∠ BAC and ab = BC, ∠ bad = ∠ DAC, ad is perpendicular to BC,
And because AE is the square line of the outer corner of ∠ BAC, so ∠ CAE = ∠ EAF (F is the point on the extension line of BA)
Therefore, CAD + CAE = 90 * means ead is right angle
Because of CE vertical AE,
So the quadrilateral adce is a rectangle
(2) Because the quadrilateral adce is rectangular, we can get ∠ ODC = ∠ OCD (o is AC, de intersection) AE parallel BC
Because AB = AC, so ∠ B = OCD, so ∠ B = ODC, so AB is parallel to de,
So quadrangle ABDE is parallelogram

Find an indefinite integral ∫ 10 ^ (2arccos x) / √ (1-x ^ 2) DX
What I calculated is - 10 ^ (2arccosx + 1) / (arccosx + 1) + C
- 10^（2arccosx) / 2In10 +C
Here are my steps
Let me first set X = cost
The original formula = ∫ 10 ^ (2arccos x) / sint dcost = - ∫ 10 ^ (2arccos x) DT
=-∫ 10^（2arccos x）d arccosx=-10^（2arccosx+1) / (arccosx +1)+C

You replace 2arccosx with S
-∫ 10^（2arccos x）d arccosx=-∫ 100^s d s=-100^s/ln100+c
I don't understand how you came from behind. Maybe you are wrong