# Line segment joining the midpoint of the hypotenuse

1
by NripChaudhuri460

2015-05-09T13:40:23+05:30

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GIVEN ;-

⇒ In Right angled triangle ABC ,

⇒ In the below figure , it is given that D is midpoint of AC.

TO PROVE :-

⇒  We need to prove that , - BD = 1 / 2 × AC

PROOF ;-

⇒In Δ ABC ,  The line L is parallel to AB.

⇒Therefore we can say that ,

ABC = DEC = 90°  { As line L is parallel to AB}

We know that D is the mid point of AC in
Δ ABC,

⇒ So  -  DE || AB [ D is the mid point of AC]

Using converse of mid point theorem   we get ,

E is the mid point of DC

⇒     Therefore                BE  =  EC    → ( Equation no . 2 )

⇒                  Now ,  In    Δ  DEB and Δ  DEC,

⇒ ∠DEB  = ∠DEC =  90º   [ from Equation  (1)]

⇒  DE      =    DE               { Common side }

⇒  BE      =    EC               [From Equation  (2)]

So by SAS congruence rule we get

Δ  DEB  ≅  Δ  DEC

DC      =    BD  [ By C.P.C.T ]

we know that D is the mid point of AC- so we get as ,

⇒

Therefore we get  as BD = 1/2 of AC

Hence proved.