Can someone help and explain and do the solution pls :(
1 answer:
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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Answer:
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