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Question

ABC is a triangle right angled at C. A line through the mid point M of hypotenuse AB and parallel to BC intersect AC at D .Show that :
CM=MA=12AB

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Solution

InΔABC

M is the midpoint of AB

and MDBC
therefore D is the midpoint of AC [line drawn through midpoint of one side of a triangle parallel to another side, bisect the third side]

AD=DC(1)

also, ADM=CDM=90o(correspondingangle)InΔADMandΔCDMAD=CD(from(1))ADM=CDM=90oDM=DM(common)ΔADMΔCDM(SAS)AM=CM(CPCT)AM=12AB(MisthemidpointofAB)CM=MA=12AB

1170110_1184848_ans_09cccf06374c458aa3d310ccafcaf890.png

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