Question

plz answer this sir and mam. for my exam. help me please

Answer 1

Let, ∠ABD=∠DBC=X; ∠ACO=∠OCB=Y; ∠CDB=Z
From ΔABC we have,
 ∠A+2∠X+2∠Y=180°
or, ∠X+∠Y=90°-∠A/2 -------------(1)
rom ΔBCD we have,
∠X+2∠Y+∠Z=180°
OR, ∠X+∠Y+∠Y+∠Z=180°
OR, ∠Y+∠Z=180°-(∠X+∠Y) 
OR, ∠Y+∠Z=180°-90°+∠A/2 [using (1)]
OR, ∠Y+∠Z=90°+∠A/2 -------------------(2)
From ΔDOC we have,
∠DOC+∠OCD+∠CDO=180°
OR, ∠DOC+∠Y+∠Z=180°
OR, ∠DOC=180°-(∠Y+∠Z)
OR, ∠DOC=180°-90°-∠A/2 [using (2)]
OR, ∠DOC=90°-∠A/2
OR, 2∠DOC=180°-∠A
OR, ∠A+2∠DOC=180°

Answer 2

Hey there !

===== Answer is here =====

Let the angle :-

angle ABD = angle DBC = x
angle ACO = angle OCB = y
angle EBD = z

From Angle ABC :-

<A+2<x+2<y = 180° ----------- 1

From angle BCD :-

<x+2<y+z =180° ------------2

By using equation 1&2

<y+<z = 90°+<A/2 ------------ 3

From angle DOC we have....

<DOC +<OCD <CDO = 180°

By using 3 equation...

<DOC =90°-A/2

2<DOC =180°-<A

<A+2 <DOC = 180°

Thank you