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Question Number 44103 by Raj Singh last updated on 21/Sep/18
Commented by Raj Singh last updated on 21/Sep/18
inthisfigurePQ∣∣BAandPR∣∣CAifPD=12findBD×CD
Answered by MrW3 last updated on 21/Sep/18
ΔDBR∼ΔDPQBDPD=BRPQΔDPR∼ΔDCQPDCD=PRCQΔBPR∼ΔPCQBRPQ=PRCQ⇒BDPD=PDCD⇒BD×CD=PD2=122=144
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