ABCD is a rhombus if angle ACB=40 then angle ADB?????
The diagonals in a rhombus are perpendicular,
So, ∠BPC=90
From triangle BPC, the sum of angles is 180 degrees.
So, ∠CBP=180-40-90=50
Since triangle ABC is isoscles, we have
AB=BC
So, ∠ACB = ∠CAB = 40
Again from triangle APB,
∠PBA = 180-40-90=50
Again, triangle ADB is isosceles,
So,
∠ADB =∠DBA =50 degrees
So ∠ADB = 50 degrees