Question

Triangle ABC is isosceles triangle. in which median ab=ac and d is mid point of side bc then prove that centroid,circumcentre, incentre and orthocentre lie in a straight line ad.​

Answer 1

For circumcentre we have to show AD is perpendicular bisector of BC.

In Δ ABD and Δ ADC,

AB = AC (given)

AD = AD (common)

BD = DC (D is midpoint of BC)

Δ ABD ≅ Δ ADC (BY SSS congruency)

⇒ ∠ ADB + ∠ ADC = 180°

⇒ AD ⊥ BC,

⇒ BD = DC

So AD is perpendicular bisector of BC>

So, the circumcentre lie on AD.

For incentre we have to show AD is bisector of ∠BAC.

Since Δ ABD ≅ Δ ADC

⇒ ∠BAD = ∠CAD ( By CPCT)

⇒ AD is the bisector of ∠BAC.

Hence, incenter lies on AD.

For orthocenter we need to prove AD is altitude corresponding to side BC.

Since Δ ABD ≅ Δ ADC

⇒ ∠ ADB + ∠ ADC = 180°

⇒ AD ⊥ BC,

⇒ AD is altitude corresponding to side BC.

For centroid we have to prove that AD is median corresponding to BC.

Since, it is given that D is the midpoint of BC. Ad is the median.

So, centroid lies on AD.

Hence Proved