Mathematics
In the adjoining quadrilateral ABCD, AB is the longest side and DC is the shortest side. Prove that :
(i) ∠C > ∠A
(ii) ∠D > ∠B
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 eq.(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 eq.(1) and (2), we get:
⇒ ∠5 + ∠3 > ∠6 + ∠8
⇒ ∠D > ∠B
Hence, proved that ∠D > ∠B.
Related Questions
In the adjoining figure, O is the centre of a circle, XY is a diameter and XZ is a chord. Prove that XY > XZ.
In the adjoining figure, PL ⊥ QR; LQ = LS and LR > LQ. Show that PR > PQ.
Can you construct a △ABC in which AB = 5 cm, BC = 4 cm and AC = 9 cm? Give reason.
In the adjoining figure, △ABC is equilateral and D is any point on AC. Prove that:
(i) BD > AD
(ii) BD > DC
