what third step could i use to prove these triangles congruent?
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In triangle JKP and triangle JMQ,
JP=JQ(given)
JM=JK(sides of a square)
angleJMQ=angleJKP(90°each)
Therefore triangle JKP is congruent to JMQ byRHS.
JP=JQ(given)
JM=JK(sides of a square)
angleJMQ=angleJKP(90°each)
Therefore triangle JKP is congruent to JMQ byRHS.
twelftholmes:
ah thanks! i can't believe i forgot about the sides of a square..
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