Mathematics
In a quadrilateral ABCD; prove that :
(i) AB + BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(iii) AB + BC + CD + DA > 2BD
Answer
Let ABCD be the quadrilateral. Join AC and BD.
(i) In △ ABC,
⇒ AB + BC > AC (Sum of two sides in a triangle is greater tha the third triangle) ……..(1)
In △ ACD,
⇒ AC + CD > DA (Sum of two sides in a triangle is greater tha the third triangle) …………(2)
Adding equations (1) and (2), we get :
⇒ AB + BC + AC + CD > AC + DA
⇒ AB + BC + CD > AC + DA - AC
⇒ AB + BC + CD > DA ……..(3)
Hence, proved that AB + BC + CD > DA.
(ii) In △ ACD,
⇒ CD + DA > AC (Sum of two sides in a triangle is greater tha the third triangle) …………(4)
Adding equations (1) and (4), we get :
⇒ AB + BC + CD + DA > AC + AC
⇒ AB + BC + CD + DA > 2AC.
Hence, proved that AB + BC + CD + DA > 2AC.
(iii) In △ ABD,
⇒ AB + DA > BD (Sum of two sides in a triangle is greater tha the third triangle) ………(5)
In △ BCD,
⇒ BC + CD > BD (Sum of two sides in a triangle is greater tha the third triangle) ………(6)
Adding equations (5) and (6), we get :
⇒ AB + DA + BC + CD > BD + BD
⇒ AB + BC + CD + DA > 2BD.
Hence, proved that AB + BC + CD + DA > 2BD.
Related Questions
In the following figure;
AC = CD; ∠BAD = 110° and ∠ACB = 74°.
Prove that : BC > CD.
From the following figure; prove that :
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC
In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that :
(i) BP > PA
(ii) BP > PC
P is any point inside the triangle ABC. Prove that : ∠BPC > ∠BAC.