Question

In quadrilateral ABCD, AB is the shortest whereas CD is the longest side. Prove AngleB>AngleD Attach figure pls

Answer 1

Proof :- 

In quad ABCD ( See the given Figure ) ,We can find two triangles - ΔBCD & ΔABD

→ Now , In ΔBCD

∠BDC > ∠ CBD{ Angles opposite to smaller side is smaller } → 1st Equation

→Now , In Δ ABD 

∠ADB > ∠ABD { Angles opposite to smaller side is smaller } → 2nd Equation

Let us now add the first two equations respectively ,

→ ∠ BDC + ∠ADB   >   ∠CBD  + ∠ ABD

→ ∠ ABC  > ∠ ADB 

⇒ Hence proved that ∠ B  > ∠ D