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Free_Kalibri [48]
3 years ago
8

Can someone help and explain and do the solution pls :(

Mathematics
1 answer:
brilliants [131]3 years ago
6 0

Consider ∆JWZ and ∆JKZ

WZ~KJ (given)

<u>/</u><u> </u><u>WZJ</u>~<u>/</u><u> </u>KJZ (given)

JZ~JZ (common)

Therefore,

∆JWZ~∆JKZ by SAS congruence rule.

JW~ZK by CPCT.

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