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
InΔABCM is the midpoint of AB
and MD∥BC
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)∴ΔADM≅ΔCDM(SAS)∴AM=CM(CPCT)∵AM=12AB(MisthemidpointofAB)∴CM=MA=12AB