In a triangle ABC, co-ordinates of A is ( 2, 5 ) and the centroid of triangle is ( -2, 1 ), let us find the co-ordinates of mid point of BC.
Answers
Answer:
For circumcentre for a triangle
It lines on the perpendicular bisector, so we have to show that AD is perpendicular bisector of the side BC.
Since ΔABD≅ΔACD (by SSS)
⇒∠ADB=∠ADC
=90
o
[∠ADB+∠ADC=180
o
]
⇒AD⊥BC
Now, AD ⊥ BC and BD = CD
⇒ AD is perpendicular bisector of BC
Hence, circumcentre lies on AD
For incentre of a triangle
It lies on the angle bisector so we have to shown that AD is bisector of ∠BAC
Since ΔABD≅ΔACD (by SSS)
⇒∠BAD=∠CAD (CPCT)
⇒ AD is bisector of triangle BAC
Hence, incentre lies on AD
For orthocentre of a triangle.
It lies on the altitude, so we have to show that AD is altitude corresponding to side BC'.
Since ΔABD≅ΔACD (by SSS)
⇒∠ADB=∠ADC
=90
o
[∠ADD+∠ADC=180
o
]
AD is perpendicular to BC' .i.e. AD is altitude corresponding to side BC.
Hence orthocentre lies on AD.
For centroid of a triangle
It lies on the median so we have to prove that AD is median corresponding to side BC.
Since, it is given that D is the mid-point of BC. AD is the median of triangle.
Hence centroid lies on AD. Hence proved.
Answer:
For circumcentre for a triangle
It lines on the perpendicular bisector, so we have to show that AD is perpendicular bisector of the side BC.
Since ΔABD≅ΔACD (by SSS)
⇒∠ADB=∠ADC
=90
o
[∠ADB+∠ADC=180
o
]
⇒AD⊥BC
Now, AD ⊥ BC and BD = CD
⇒ AD is perpendicular bisector of BC
Hence, circumcentre lies on AD