Question

♱Qᴜᴇꜱᴛɪᴏɴ♰

♱Qᴜᴇꜱᴛɪᴏɴ♰
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ꜱʜᴏᴡ ᴛʜᴀᴛ ɪꜰ ᴅɪᴀɢᴏɴᴀʟꜱ ᴏꜰ ꜱQᴜᴀʀᴇ ᴀʀᴇ ᴇQᴜᴀʟ ᴀɴᴅ ʙɪꜱᴇᴄᴛ ᴇᴀᴄʜ ᴏᴛʜᴇʀ ᴀᴛ ʀɪɢʜᴛ ᴀɴɢʟᴇꜱ, ᴛʜᴇɴ ɪᴛ ɪꜱ ᴀ ꜱQᴜᴀʀᴇ

♱Qᴜᴇꜱᴛɪᴏɴ♰
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ꜱʜᴏᴡ ᴛʜᴀᴛ ɪꜰ ᴅɪᴀɢᴏɴᴀʟꜱ ᴏꜰ ꜱQᴜᴀʀᴇ ᴀʀᴇ ᴇQᴜᴀʟ ᴀɴᴅ ʙɪꜱᴇᴄᴛ ᴇᴀᴄʜ ᴏᴛʜᴇʀ ᴀᴛ ʀɪɢʜᴛ ᴀɴɢʟᴇꜱ, ᴛʜᴇɴ ɪᴛ ɪꜱ ᴀ ꜱQᴜᴀʀᴇ

♱Qᴜᴇꜱᴛɪᴏɴ♰
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ꜱʜᴏᴡ ᴛʜᴀᴛ ɪꜰ ᴅɪᴀɢᴏɴᴀʟꜱ ᴏꜰ ꜱQᴜᴀʀᴇ ᴀʀᴇ ᴇQᴜᴀʟ ᴀɴᴅ ʙɪꜱᴇᴄᴛ ᴇᴀᴄʜ ᴏᴛʜᴇʀ ᴀᴛ ʀɪɢʜᴛ ᴀɴɢʟᴇꜱ, ᴛʜᴇɴ ɪᴛ ɪꜱ ᴀ ꜱQᴜᴀʀᴇ



♱Qᴜᴇꜱᴛɪᴏɴ♰
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ꜱʜᴏᴡ ᴛʜᴀᴛ ɪꜰ ᴅɪᴀɢᴏɴᴀʟꜱ ᴏꜰ ꜱQᴜᴀʀᴇ ᴀʀᴇ ᴇQᴜᴀʟ ᴀɴᴅ ʙɪꜱᴇᴄᴛ ᴇᴀᴄʜ ᴏᴛʜᴇʀ ᴀᴛ ʀɪɢʜᴛ ᴀɴɢʟᴇꜱ, ᴛʜᴇɴ ɪᴛ ɪꜱ ᴀ ꜱQᴜᴀʀᴇ​

Answer 1

Answer:

Question: Show that the diagonals of a a square are equal and bisect each other at right angles.

To prove: (i) AC = BD

(ii) OA=OC OB = OD

iii) angle 1 = angle 2 = angle 3 = angle 4 = 90°

In ∆ABD and ∆ABC

AB = AB (common)

AD = BC (sides of square)

Angle A = Angle B (90°)

∆ABD ~ ∆ABC (by SAS)

AC = BD (by cpct)

In ∆DOC and ∆AOB

DC = AB (sides of square)

Angle 2 = Angle 4 (vertically opposite angles)

Angle 5 = Angle 6 (AIA)

∆DOC ~ ∆AOB (by ASA)

OA = OC

OD = OB (by cpct)

In ∆AOD and ∆COD

OD = OD (common)

AD = CD (sides of square)

OA = OC (proved)

∆AOD ~ ∆COD ( by SSS)

Angle 1 = Angle 2 (by cpct)

Similarly, Angle 3 = Angle 4

Angle 1 + Angle 2 = 180°

2angle 1 = 180°

Angle 1 = 180° /2 = 90°

Angle 1 = 90°

Angle 1 = Angle 2= 90°

Angle 2= Angle 4 = 90°

Angle 1 = Angle 3 = 90°