AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.
Solution:
Given: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD
To prove: ∠A > ∠C and ∠B > ∠D
Let's join vertex A to C as shown below.
In the above ΔABC,
AB < BC (given AB is the smallest side of quadrilateral ABCD)
∠ACB < ∠BAC (Angle opposite to the smaller side is smaller) ... (1)
In ΔADC,
AD < CD (given CD is the largest side of quadrilateral ABCD)
∠ACD < ∠CAD (Angle opposite to the smaller side is smaller) ... (2)
On adding Equations (1) and (2), we obtain
∠ACB + ∠ACD < ∠BAC + ∠CAD
∠C < ∠A
This means, ∠A > ∠C
Hence Proved.
Let's now join BD.
In ΔABD,
AB < AD (Given AB is the smallest side of quadrilateral ABCD)
∠ADB < ∠ABD (Angle opposite to the smaller side is smaller) ...(3)
In ΔBDC,
BC < CD (Given CD is the largest side of quadrilateral ABCD)
∠BDC < ∠CBD (Angle opposite to the smaller side is smaller) ... (4)
On adding equations (3) and (4), we obtain
∠ADB + ∠BDC < ∠ABD + ∠CBD
∠D < ∠B
This means, ∠B > ∠D
Hence, proved.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 7
Video Solution:
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠A > ∠C and ∠B > ∠D.
NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.4 Question 4
Summary:
If AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD, then we have proved that ∠A >∠C and ∠B >∠D
☛ Related Questions:
- Show that in a right-angled triangle, the hypotenuse is the longest side.
- In Fig. 7.48, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
- In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
- In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.