Question

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure 7.50) . Show that Angle A is greater than Angle C and Angle B is greater than Angle D.​

Answer 1

Answer:

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Answer 2

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Given

• AB is the smallest side
• CD is the longest side

To Prove

• ∠A > ∠C
• ∠B > ∠D

Construction

• Join A and C
• Join B and D

Proof

In △ABC

BC > AB [ AB is smallest side ]

∠1 > ∠4───eq.1 [ Angle opposite to greater side is greater ]

In △ACD

CD > AD [ CD is longest side ]

∠2 > ∠3───eq.2 [ Angle opposite to greater side is greater ]

By adding eq. 1 and 2

∠ 1 + ∠2 > ∠4 + ∠3

∠A > ∠C

In △ABD

AB > AD [ AB is smallest side ]

∠5 > ∠8───eq.3 [ Angle opposite to greater side is greater ]

In △BDC

CD > BC [ CD is longest side ]

∠6 > ∠7───eq.4 [ Angle opposite to greater side is greater ]

By adding eq. 3 and 4

∠ 5 + ∠6 > ∠8 + ∠7

∠B > ∠D

Hence Proved✔