Suppose ABC is a triangle and D is the midpoint of BC. Assume that the perpendiculars from D to AB and AC are of equal length. Prove that ABC is isosceles.
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Given :
In ∆ABC , D is the midpoint of BC.
The perpendiculars from D to AB
and AC are equal.
To prove :
∆ABC is isosceles .
Proof :
In right triangles DEC and DFB ,
Hypotenuse DC = Hypotenuse DB
[ D is the midpoint of BC ]
side DE = side DF [ given ]
Therefore ,
∆DEC is congruent to ∆DFB
[ RHS congruence rule ]
<DCE = <DBF
=> <BCA = <CBA
AB = AC
[ Sides opposite to equal angles
of a triangle are equal ]
••••
In ∆ABC , D is the midpoint of BC.
The perpendiculars from D to AB
and AC are equal.
To prove :
∆ABC is isosceles .
Proof :
In right triangles DEC and DFB ,
Hypotenuse DC = Hypotenuse DB
[ D is the midpoint of BC ]
side DE = side DF [ given ]
Therefore ,
∆DEC is congruent to ∆DFB
[ RHS congruence rule ]
<DCE = <DBF
=> <BCA = <CBA
AB = AC
[ Sides opposite to equal angles
of a triangle are equal ]
••••
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Answer:
Given :
In ∆ABC , D is the midpoint of BC.
The perpendiculars from D to AB
and AC are equal.
To prove :
∆ABC is isosceles .
Proof :
In right triangles DEC and DFB ,
Hypotenuse DC = Hypotenuse DB
[ D is the midpoint of BC ]
side DE = side DF [ given ]
Therefore ,
∆DEC is congruent to ∆DFB
[ RHS congruence rule ]
<DCE = <DBF
=> <BCA = <CBA
AB = AC
[ Sides opposite to equal angles
of a triangle are equal
Step-by-step explanation:
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