Q.No.8. AB and CD are respectively the smallest and longest sides of quadrilateral<br />ABCD. Show that ZA > ZC and B > ZD<br />noint S if the bisector of
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Q.No.8. AB and CD are respectively the smallest and longest sides of quadrilateral<br />ABCD. Show that ZA > ZC and B > ZD<br />noint S if the bisector of
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Let ABCD be a quadrilateral such that AB is it's smallest side and CD is it largest side.
Now join AC and BD
In ΔABC ,we have
BC>AB
⇒∠8>∠3 ...(1) {∵ angle opposite to longer side is greater.}
Since CD is the longest side of quadrilateral ABCD
In ΔACD, we have
CD>AD
⇒∠7>∠4 ...(2)
Adding (1) and(2) we get
∠8+∠7>∠3+∠4
⇒∠A>∠C
Now in ΔABD, we have
AD>AB {∵AB is the shortest side}
⇒∠1>∠6 ...(3)
In ΔBCD,we have
CD>BC
⇒∠2>∠5 ....(4)
Adding (3) and (4)
∠1+∠2>∠5+∠6
⇒∠B>∠D
Hence proved.