Help Irma wants to prove that the side-angle-side(SAS) criterion is sufficient to show thatALMNAXYZ.LNZX++MYIrma starts by translating ALMN to get AL'M'N',where N' = Z and M' = Y. Then she reflectsAL'M'N' across YZ to get AL "M"N".Irma says that YX and M"L" coincide. How canIrma justify this statement? XIrma starts by translating ALMN to get AL'M'N',where N' = Z and M' = Y. Then she reflectsAL'M'N' across YZ to get AL"M"N".Irma says that YX' and M"L" coincide. How canIrma justify this statement?Translation and reflection preserveangle measure, so mZL"N"M" =mZXZY.Translation and reflection preserveangle measure, so mZL"M"N":mZXYZ.=Translation and reflection preservedistance, so L"M" = XY.Translation and reflection preservedistance, so M"N" = YZ.

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X rma starts by translating ALMN to get AL'M'N', where N' = Z and M' = Y. Then she reflects L'M'N' across YZ to get AL "M"N". rma says that YX and M"L" coincide. How can rma justify this statement? Translation and reflection preserve angle measure, so mZL"N"M" : mZXZY. = Translation and reflection preserve angle measure, so mZL"M"N" mZXYZ. = Translation and reflection preserve distance, so L"M" = XY. Translation and reflection preserve distance, so M"N" = YZ.