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Byju's Answer
Standard IX
Mathematics
Converse of Pythagoras Theorem by Contradiction
Triangle ABC ...
Question
Triangle ABC is right angled at
C
. A line through the mid-point
M
of hypotenuse
A
B
and parallel to
B
C
, intersects AC at D.
Show that:
C
M
=
M
A
=
1
2
A
B
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Solution
Consider a right angle
Δ
A
B
C
right angled at C.
⇒
∠
C
=
90
∘
And
M
is the mid-point of
A
B
.
and DM
∥
BC
⇒
D
is the midpoint of AC [By converse of midpoint theorem]
In
Δ
A
D
M
and
Δ
C
D
M
,
D
M
=
M
D
[Common]
A
D
=
C
D
[Since, D is mid-point of AC]
∠
A
D
M
=
∠
C
D
M
[Since
D
M
∥
B
C
,
∠
A
D
M
=
90
∘
]
Δ
A
D
M
≅
Δ
C
D
M
[SAS congruency]
C
M
=
A
M
……(i) [CPCT]
A
M
=
B
M
=
1
2
A
B
..........(ii)
From Equations (i) and (ii) , we get
C
M
=
A
M
=
1
2
A
B
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Q.
ABC is a triangle right angled at C. A line through the mid point M of hypotenuse AB and parallel to BC intersect AC at D .Show that :
C
M
=
M
A
=
1
2
A
B
Q.
A
B
C
is a triangle right angled at
C
.
A
line through the mid-point
M
of hypotenuse
A
B
and parallel to
B
C
intersects
A
C
at
D
.
Show that
(
i
)
D
is the mid-point of
A
C
(
i
i
)
M
D
⊥
A
C
(
i
i
i
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C
M
=
M
A
=
1
2
A
B
Q.
ABC is a triangle right angled at C. A line through the mid point M of the hypotenuse AB and parallel to BC intersects AC at D. Show that,
(i) D is the mid-point of AC.
(ii)
M
D
⊥
A
C
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M
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B
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Converse of Pythagoras Theorem by Contradiction
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