Question

ABCD is a quadrilateral.
Is AB+BC+CD+DS>AC+BD?

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Answer 1

ᴀɴꜱᴡᴇʀ :

ᴀʙᴄᴅ ɪꜱ ᴀ Qᴜᴀᴅʀɪʟᴀᴛᴇʀᴀʟ ᴀɴᴅ ᴀᴄ, ᴀɴᴅ ʙᴅ ᴀʀᴇ ᴛʜᴇ ᴅɪᴀɢᴏɴᴀʟꜱ.

ꜱᴜᴍ ᴏꜰ ᴛʜᴇ ᴛᴡᴏ ꜱɪᴅᴇꜱ ᴏꜰ ᴀ ᴛʀɪᴀɴɢʟᴇ ɪꜱ ɢʀᴇᴀᴛᴇʀ ᴛʜᴀɴ ᴛʜᴇ ᴛʜɪʀᴅ ꜱɪᴅᴇ.

ꜱᴏ, ᴄᴏɴꜱɪᴅᴇʀɪɴɢ ᴛʜᴇ ᴛʀɪᴀɴɢʟᴇ ᴀʙᴄ, ʙᴄᴅ, ᴄᴀᴅ ᴀɴᴅ ʙᴀᴅ, ᴡᴇ ɢᴇᴛ

ᴀʙ + ʙᴄ > ᴀᴄ

ᴄᴅ + ᴀᴅ > ᴀᴄ

ᴀʙ + ᴀᴅ > ʙᴅ

ʙᴄ + ᴄᴅ > ʙᴅ

ᴀᴅᴅɪɴɢ ᴀʟʟ ᴛʜᴇ ᴀʙᴏᴠᴇ ᴇQᴜᴀᴛɪᴏɴꜱ,

2(ᴀʙ + ʙᴄ + ᴄᴀ + ᴀᴅ) > 2(ᴀᴄ + ʙᴅ)

⇒ 2(ᴀʙ + ʙᴄ + ᴄᴀ + ᴀᴅ) > 2(ᴀᴄ + ʙᴅ)

⇒ (ᴀʙ + ʙᴄ + ᴄᴀ + ᴀᴅ) > (ᴀᴄ + ʙᴅ)

ʜᴇɴᴄᴇ, ᴘʀᴏᴠᴇᴅ✔️✔️

Answer 2

Answer:

ABCD is a quadrilateral and AC, and BD are the diagonals.

Sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

⇒ (AB + BC + CA + AD) > (AC + BD)

HENCE, PROVED

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