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

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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**

→

∠BDC > ∠ CBD

→

∠ADB > ∠ABD

→ ∠ ABC

⇒ Hence proved that