Sol plz

9+17=26

Let x=ad=bd
Y=cd
9=2(x^2+y^2)
X^2+y^2=9/2

1/√2=3+y^2-x^2/2√3y

√6y=3+y^2-x^2.........@
√6y=3+y^2-(9/2-y^2)
√6y=-3/2+2y^2
2√6y=-3+4y^2
4y^2-2√6y-3=0
Y=2√6+-√24+48/8
=2√6+-6√2/8
=2√6+6√2/8=√6+3√2/4=√3+3/2√2

From equation@
Y^2-x^2=√6y-3
=√6(3+√3)-6√2/2✓2
=3√6-3√2/2√2
=3√3-3/2
Cosx=6+3√3-3/2)/2√6y
=9+3√3/4√6y
=3√3+3/4√2y
=3(√3+1)*2√2/4√6(√3+1)
=6√2/4√6=3√2/2√6=3/2√3=√3/2
X=30

Let AD= BD = x

Sin45/x = sinθ/√3
sinθ*x= √3/√2
Now in triangle BCD
Sin(180-θ)/√6 = sinx/x
Sinθ/√6 = sinx/x
Sinθ*x=sinx*√6
Sinx=√3/√2*√6= 1/2
X= 30°

@Cylus12 sol plz sir

Any obe ?