q9 answer please very urgent
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ʜᴇʟʟᴏ ᴍᴀᴛᴇ!
(i) In ∆ DCN and ∆ ABP, we have
DC = BA ( equal sides of rectangle )
/_ DNC = /_ BPA ( 90° each )
/_ DCN = /_ BAP ( Alternate Angles )
Hence, both ∆s are congruent by ASA property.
Hence proved
(ii) In ∆ DNA and ∆ BPC, we have
AD = BC ( equal sides of rectangle )
/_DNA = /_ BPC ( 90° each )
/_ DAN = /_ BCP ( Alternate angles )
Hence both ∆s are congruent by ASA peoperty.
Therefore ; AN = CP ( c.p.c.t )
Hence proved
Hope it helps
(i) In ∆ DCN and ∆ ABP, we have
DC = BA ( equal sides of rectangle )
/_ DNC = /_ BPA ( 90° each )
/_ DCN = /_ BAP ( Alternate Angles )
Hence, both ∆s are congruent by ASA property.
Hence proved
(ii) In ∆ DNA and ∆ BPC, we have
AD = BC ( equal sides of rectangle )
/_DNA = /_ BPC ( 90° each )
/_ DAN = /_ BCP ( Alternate angles )
Hence both ∆s are congruent by ASA peoperty.
Therefore ; AN = CP ( c.p.c.t )
Hence proved
Hope it helps
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