Mathematics
In quadrilateral ABCD, side AB is the longest and side DC is the shortest. Prove that :
(i) ∠C > ∠A
(ii) ∠D > ∠B
Triangles
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Answer
(i) Given,
In quadrilateral ABCD,
AB is the longest sides and DC is the shortest side.
Join BD and AC.
In △ ABC,
⇒ AB > BC
∴ ∠1 > ∠2 …….(1) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]
In △ ADC,
⇒ AD > DC
∴ ∠7 > ∠4 …….(2) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]
Adding equations (1) and (2), we get :
⇒ ∠1 + ∠7 > ∠2 + ∠4
⇒ ∠C > ∠A.
Hence, proved that ∠C > ∠A.
(ii) In △ ABD,
⇒ AB > AD
∴ ∠5 > ∠6 …….(1) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]
In △ BDC,
⇒ BC > CD
∴ ∠3 > ∠8 …….(2) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]
Adding equations (1) and (2), we get :
⇒ ∠5 + ∠3 > ∠6 + ∠8
⇒ ∠D > ∠B.
Hence, proved that ∠D > ∠B.
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