Question -
Answer -
Letus join AC.
InΔABC,
AB< BC (AB is the smallest side of quadrilateral ABCD)
∴ ∠2 < ∠1(Angle opposite to the smaller side is smaller) … (1)
InΔADC,
AD< CD (CD is the largest side of quadrilateral ABCD)
∴ ∠4 < ∠3(Angle opposite to the smaller side is smaller) … (2)
Onadding equations (1) and (2), we obtain
∠2 + ∠4 < ∠1+ ∠3
⇒ ∠C < ∠A
⇒ ∠A > ∠C
Letus join BD.
InΔABD,
AB< AD (AB is the smallest side of quadrilateral ABCD)
∴ ∠8 < ∠5(Angle opposite to the smaller side is smaller) … (3)
InΔBDC,
BC< CD (CD is the largest side of quadrilateral ABCD)
∴ ∠7 < ∠6(Angle opposite to the smaller side is smaller) … (4)
Onadding equations (3) and (4), we obtain
∠8 + ∠7 < ∠5+ ∠6
⇒ ∠D < ∠B
⇒ ∠B> ∠D