Geo help

Angle C  =  ACB      Angle D  = ADB

BAD  = ADB         ABC  = ACB

DBA  + BAC   =  90    ⇒   DBA  = 90 - BAC 

ACB + DBC  = 90    ⇒      DBC =  90 - ACB 

DAC + ADB  = 90   ⇒       DAC  = 90 - ADB       

DAC + BAC  = BAD                      DBA + DBC = ABC

DAC + BAC  = ADB                      DBA + DBC =  ACB

[90 - ADB] + BAC  = ADB             DBA + [90 - ACB]  = ACB

90 + BAC = 2ADB   (1)                  90 + DBA  = 2ACB

                                                      90 + [90 - BAC]  = 2ACB

                                                     180  - BAC  = 2ACB

                                                      BAC  = 180 - 2ACB   (2)

Sub  (2)  into (1)

90 + [180 - 2ACB]  = 2ADB

270 =  2ACB + 2ADB

135  = ACB + ADB

135  =  ∠C +  ∠ D

Triangles ABC and ABD are isosceles with

$AB=AC=BD$,
and$\overline{BD}$ intersects$\overline{AC}$ at $E$.
If $\overline{BD}\perp\overline{AC}$, then what is the value of $\angle C+\angle D$?

$\angle D = BDA = DAB \\ \angle C = BCA = CBA \\ \angle \alpha = DBA\\ \angle \beta = BAC \\ \angle \alpha +\angle \beta = 90^{\circ}$

$\begin{array}{|rcll|} \hline \angle \alpha &=& 180^{\circ} - 2\times \angle D \\ \angle \beta&=& 180^{\circ} - 2\times \angle C \\ \angle \alpha +\angle \beta &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \quad & | \quad \angle \alpha +\angle \beta = 90^{\circ}\\ 90^{\circ} &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \\ 90^{\circ} &=& 360^{\circ} - 2\times (\angle C + \angle D) \\ 2\times(\angle C + \angle D) &=& 360^{\circ} - 90^{\circ} \\ 2\times(\angle C + \angle D) &=& 270^{\circ} \\ \mathbf{ \angle C + \angle D } & \mathbf{=}& \mathbf{135^{\circ}} \\ \hline \end{array}$