Question

ABCD are respectively the smallest and the longest sides of a quadrilateral ABCD show that angle A is greater than angle c and angle B geater than angle angleD

Answer 1

Heya!!

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

Hope this helps you ☺☺