and
are two isosceles triangles on the same base BC and vertices A and D are on the
same side of BC (see Figure). If AD is extended to intersect BC at P, show that
(i)
(ii)
(iii) AP bisects
as well as
.
(iv) AP is the perpendicular bisector of BC.
Solution (i)
In
and
AB=AC (
is isosceles)
BD=CD (
is isosceles)
AD=AD (Common)
Therefore, by SSS congruence rule,
Solution (ii)
In
and
, we have
AB=AC (
is isosceles)
(Corresponding parts of congruent triangles,
and
)
AP=AP (Common)
Therefore, by SAS congruence rule,
Solution (iii)
In
, we have
BD=CD (
is isosceles)
(Corresponding parts of congruent triangles,
and
) (1)
and
also make a linear pair.
Therefore,
(2)
From (1) and (2), we can say that
DP=DP (Common)
Therefore, by RHS congruence rule,
It means that
(Corresponding parts of congruent triangles) (3)
We have already proved above that
It means that
(Corresponding parts of congruent triangles) (4)
From (3) and (4), we can say that
AD bisects
as well as
.
Solution (iv)
We have proved above in solution (iii) that
(5)
Also, BP=CP (Corresponding parts of congruent triangles,
) (6)
From (5) and (6), we can say that AP is the perpendicular bisector of BC.
Ankit kumar says
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