AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Figure). Show that A > C and B > D.
Triangles
1
Posts
2
Users
0
Likes
156
Views
0
12/07/2021 12:26 pm
Topic starter
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Figure). Show that A > C and B > D.
Answer
Add a comment
Add a comment
1 Answer
0
12/07/2021 12:28 pm
In ΔABD, we see that
AB < AD < BD
So, ADB < ABD ...... (i) (Since angle opposite to longer side is always larger)
Now, in ΔBCD,
BC < DC < BD
Hence, it can be concluded that
BDC < CBD — (ii)
Now, by adding equation (i) and equation (ii) we get,
ADB + BDC < ABD + CBD
ADC < ABC
B > D
Similarly, In triangle ABC,
ACB < BAC — (iii) (Since the angle opposite to the longer side is always larger)
Now, In ΔADC,
DCA < DAC — (iv)
By adding equation (iii) and equation (iv) we get,
ACB + DCA < BAC+DAC
⇒ BCD < BAD
∴ A > C
Add a comment
Add a comment
Forum Jump:
Related Topics

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
2 years ago

In Figure, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.
2 years ago

In Figure, B < A and C < D. Show that AD < BC.
2 years ago

In Figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.
2 years ago

Show that in a rightangled triangle, the hypotenuse is the longest side.
2 years ago
Forum Information
 321 Forums
 27.3 K Topics
 53.8 K Posts
 1 Online
 12.4 K Members
Our newest member: Stripchat
Forum Icons:
Forum contains no unread posts
Forum contains unread posts
Topic Icons:
Not Replied
Replied
Active
Hot
Sticky
Unapproved
Solved
Private
Closed