XYZ is a triangle which is right angled at Y.A line is drawn through the midpoint m of hypotenuse XZ and parallel to YZ to intersect Y at N.show that N is the midpoint if XYand YM=XM=1/2XZ

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In XYZ, M is the mid point of XZ and MNZY, soN is the mid point of XY   By converse of mid point theoremXN = YN    ...1Since, MNZY and NY is a transversal, thenMNY + ZYN = 180°  consecutive interior angles on the same side of transversal are supplementaryMNY + 90° = 180°MNY = 90°now, MNY + XNM = 180°   Linear pair90°+XNM = 180°XNM =90°In XNM and YNM   XN = YN   Using 1XNM = YNM   90° EachNM = NM   CommonXNM  YNM  SASXM = MY  CPCTNow, XM = 12XZ  As, M is the mid point of XZso, YM = XM = 12XZ

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  • 8
Using convers of Mid Point theorem:
M is midpt of XZ (given)
NM || YZ ( given)
So N will be a midpt of XY.

Now in triangle XNM and YNM,
NM = NM (common)
XN = YN ( N is midpt)
​angle XNM = angle YNM ( NM || YZ, so ​​angle XNM = angle YNM , corresponding angles.)
Hence by SAS these triangles are congruent.
CPCT- XM = MY
XM = MY , but m is midpt of XZ so XM = MZ
Thus, XM = MY = 1/2 XZ
 
  • 2
since M is midpoint of XZ and MN is parallel to YZ.
so,N is midpoint of XY.
now,so,so, MN is perpendicular to XY.
In triangles XNM amd MNY
XN=NY (proved above)
and MN=MN (common)
so,XNM congruent to MNY.
so,XM=YM (By CPCT)
and XM = 1/2 XZ
Therefore,XM=YM=1/2 XZ
  • 4
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