AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that angle A is greater than angle C and angle B is greater than angle D.

In ΔABC, we have

BC > AB

⇒ ∠BAC > ∠BCA ...(1)

Similarly, in ΔACD, we have

CD > AD

⇒ ∠CAD > ∠ACD ...(2)

On adding (1) and (2), we get

∠BAC + ∠CAD > ∠BCA + ∠ACD

⇒∠BAD > ∠BCD

⇒ ∠A > ∠C

Now, in Δ ABD, we have

AD > AB

⇒ ∠ABD > ∠ADB ...(3) [ AB is the smallest side]

Similarly, in Δ BCD, we have

CD > BC ...(4) [CD is the largest side]

⇒ ∠DBC > ∠BDC

On adding (3) and (4), we get

∠ABD + ∠DBC > ∠ADB + ∠BDC

∠ABC > ∠ADC

⇒∠B > ∠D

Hence, it is proved that ∠A > ∠C and ∠B > ∠D

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