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


So, ∠ACB = ∠CAB = 40

Again from triangle APB, 

∠PBA = 180-40-90=50

Again, triangle ADB is isosceles,


∠ADB =∠DBA =50 degrees

So ∠ADB = 50 degrees

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