Prove that "A diagonal of a 11 gm divides it intotwo congrent triangle".
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Answered by
2
Step-by-step explanation:
opposite side of a parallelogram are equal and parallel.
In ||gm ABCD,
AB = CD
AC = BD
Let, suppose BC is a diagonal.
opposite angles are equal in ||gm,
CAB = CDB
ACD = ABD
Now, in triangle CAB and triangle BDC,
AB = CD
AC = BD
BC = BC ( Common side )
this implies,
triangle CAB =~ BDC by SSS property.
HENCE PROVED.
Answered by
4
Let PQRS be the required ||gm. Let us construct a diagonal AC. Now, we have to prove that the two triangles formed are congruent.
In △ACD and △ACB,
- AD = CB {Opposite sides of a ||gm are same}
- AC = CA {Common side}
- AB = CD {Opposite sides of a ||gm are same and ||}.
⇒△ACD ≅ △ACB (by S-S-S congruence criteria).
Hence, proved.
More:-
A given quadrilateral is a ||gm if:-
- Opposite sides are parallel.
- Opposite sides are same.
- Opposite angles are same.
- The diagonals bisect each other.
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