Alternate (trig/algebra) solution ... drop a perpendicular from A to CB extended which it meets at point P. We have two well known right-angled triangles, 45, 45 (1,1,√2)) and 30, 60 (1,√3,2) with common side AP. Let AP = s for some length s (we could make it 1 without loss of generality). let BD=DC=ds let PB=as looking at the 30, 60, 90 triangle APC PC=s√3=as+2ds => a=√3-2d ... (i) looking at the 45, 45, 90 triangle APD PA=s=PD=as+ds => d=1-a sub in ...(i) for d a=√3-2(1-a) => a=2-√3 let angle PAB = y tan y = as/s = a = 2-√3 x+y=45 degrees tan(x+y)=1=(tan x + tan y)/(1 - tan x tan y) 1=(tan x + 2-√3)/(1-tan x (2-√3)) multiply by denominator (1-tan x (2-√3) (1-tan x (2-√3))=(tan x + 2-√3) -(2-√3)tan x - tan x = -1 + 2-√3 change signs and simplify (3-√3)tan x = (√3 - 1) note that √3(√3-1)=(3-√3) √3 tan x = 1 tan x =1/√3 so x = 30 degrees

Define a as the length of BD and DC. Run a line segment from D to a point on AC that gives the new line segment a length of a. Call the endpoint of this new line segment E. Because DE=a, triangle EDC is equilateral and thus angle DEC is 30 degrees. Now note that angle EDC is 180-30-30=120 degrees. We can now find angle ADE as 180-45-120=15 degrees. Furthermore, angle DAE=180-120-15-30=15. Thus we find that triangle ADE is equilateral, and thus line segment AE=a. Now draw a line segment BE between points B and E. We know that BD=a and DE=a, therefore triangle DBE is equilateral. Furthermore, angle BDE=45+15=60, thus the other two angles, which must be equal because of the triangle being equilateral, are also both 60 degrees. Thus all 3 sides of triangle BED are equal to a, including segment BE. Now look at triangle ABE. We know AE=a and BE=a, therefore angle EAB and ABE are equal. Furthermore, we know angle BEC=30+60=90 degrees. Therefore angle BEA=90 degrees. Consequently the other two angles of the equilateral triangle ABE are both 45 degrees. Hence angle BAE=45. BAE=15+x therefore x=45-15=30.

@raqqafeller152

17 күн бұрын

sorry to be nitpicking but triangles EDC and ADE are isosceles, not equilateral. but i understood your solution anyways. 👍