In ΔABC, AB = AC. D, E and F are respectively the mid-points of sides BC, AB and AC. Show that line segment AD is perpendicular to the line segment EF and is bisected by it.
Given: In ΔABC, D, E, F are mid-points of BC, AB and AC.
Also, AB = AC
To Prove: AM = MD and ΔAMF = 90°
Construction: Join ED and FD.
Proof: In ΔABC, AB = AC ⇒
AB = AC ⇒ AE = AF...(i)
Now in the ΔABC, E is the mid-point of AB and F is the mid-point of AC.
... EF =BC and EF ïï BC
In ΔABC, E is the mid-point of AB and D is the mid-point of BC.
... ED =AC and ED ïï AC ⇒ ED = AF and ED ïï AF
Similarly, DF = AE and DF || AE
... AFDE is a parallelogram
In a parallelogram, opposite sides are equal
Also,
AE = AF ... (AB = AC)
... AFDE is a rhombus.
The diagonals of a rhombus bisect each other perpendicularly.
... AM = MD and EM = MF
and AD ⊥ EF
Also, AB = AC
To Prove: AM = MD and ΔAMF = 90°
Construction: Join ED and FD.
Proof: In ΔABC, AB = AC ⇒
AB = AC ⇒ AE = AF...(i)Now in the ΔABC, E is the mid-point of AB and F is the mid-point of AC.
... EF =
BC and EF ïï BCIn ΔABC, E is the mid-point of AB and D is the mid-point of BC.
... ED =
AC and ED ïï AC ⇒ ED = AF and ED ïï AFSimilarly, DF = AE and DF || AE
... AFDE is a parallelogram
In a parallelogram, opposite sides are equal
Also,
AE = AF ... (
AB = AC)... AFDE is a rhombus.
The diagonals of a rhombus bisect each other perpendicularly.
... AM = MD and EM = MF
and AD ⊥ EF
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