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ABC is a triangle right-angled at C. A l...

ABC is a triangle right-angled at C. A line through the midpoint P of hypotenuse AB and parallel to BC intersects AC at M. Show that
`"(i) "` M is the midpoint of `AC,`
` "(ii) " MP bot AC ,`
` "(iii) " CP = AP=(1)/(2)AB. `

Text Solution

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In `triangle ACB`, we have
P is the midpoint of AB and PM||BC.
`therefore ` M is the midpoint of AC.
[by converse of midpoint theorem].
Now, PM||BC.
`therefore angle PMC + angle BCM=180^(@) " " ("co-interior "angle )`
`rArr angle PMC + 90^(@) =180^(@) rArr angle PMC=90^(@).`
Thus, `MP bot AC.`
Join PC.
In `triangle PMA` and PMC, we have:
`MA= MC " " [because " M is the midpoint of AC"]`
`angle PMA = angle PMC " " ("each equal to "90^(@))`
`PM=PM " " `(common)
`therefore triangle PMA ~=triangle PMC " " `(SAS-criterion).
And so, `AP=CP " " `(c.p.c.t.).
Now, P is the midpoint of AB.
`therefore CP=AP=(1)/(2)AB.`
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