Question

ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and Parallel to BC intersects AC at D. Show that
(i) D is the midpoint of AC
(ii) MD⊥AC
(iii) CM = MA =1/2AB.

Answer 1

1.theorm will be applied-A line that bisect the first side(AB), parallel to second side(BC) bisect the third side(AD)

Answer 2

Given that in ∆ABC ,

<C = 90° .

M is the midpoint of AB .

i ) If ' D ' is the midpoint of AC ,

The proof is tribal .

Let us suppose D is not the midpoint

of AC .

Then there exists D' such that

AD' = D'C

D'M is a line parallel to BC through M.

Also DM is a line parallel to BC through M.

There exist two lines parallel to same

line through a point M .

This is a contradiction .

There exists only one line parallel to

a given line through a point not on

the line .

Therefore ,

D' must coincides with D .

D is the mid point of AC .

ii ) From ( i ) ,

DM // BC

<ADM = <ACB = 90°

[ Corresponding angles ]

=> MD perpendicular to AC .

iii ) In ∆ADM and ∆CDM

AD = CD

[ D is the midpoint from ( i ) ]

<ADM = <MDC = 90°

DM = DM ( common side )

∆ADM congruent to ∆CDM

[ SAS Congruence ]

=> CM = MA [ CPCT ]

CM = ( 1/2 ) AB

[ M is the midpoint of AB ]

Therefore ,,

CM = MA = AB/2

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