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ques.no. 3, 6 and 8.
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ans. Given: D,E,F are the mid points of AB,BC,CA
To Prove: DE = EF
Proof: AB = AC (isosceles triangle)
∠B = ∠C (angle opp to equal sides)
¹/₂ AB = ¹/₂ AC
so,DB = CF
In ΔADF and ΔCEF
DB = CF (proved above)
BE = CE (E is the mid point of BC)
∠B = ∠C (proved above)
∴ ΔADF = ΔCEF (SAS)
DE = FE (CPCT)
Hence, Proved
que 6
ans. Let a square ABCD in which L,M,N&O are the midpoints .
in triangle AML and triangle CNO
AM = CN ( AB = DC and M and O are the midpoints )
AL = CM ( AD = BC and L and N are the midpoints )
angle MAL = angle NCO ( all angles of a square = 90 degree )
by AAS critaria
triangle AML CONGRUENT to triangle CNO
therefore ML = ON ( CPCT )
similarly in triangle MBN CONGRUENT to LDO and
AND triangle AML is CONGRUENT to triangle
now ,
in Triangle AML ,
angle AML = angle ALM ( AM = AL )
= 45 degree
similarly in triangle LDO
angle DLO = 45 degree
there fore ,
angle MLO = 90 degree
by the properties of SQUARE
all sides are equal and angles are 90 degree
hope you understand