Question

Prove that in an equilateral triangle incentre circumcentre centroid and orthocentre are coincides

Answer 1

Answer:

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Step-by-step explanation:

Proof:

In ΔABD and ΔACD

AB = AC [Sides of equilateral triangle]

AD = AD [Common side]

BD = CD [As, AD is median to BC]

⇒ ΔABD ≅ ΔACD [By SSS property of congruent triangles]

Now, As corresponding parts of congruent triangles are equal [CPCT], we have

∠ADB = ∠ADC

Also,

∠ADB + ∠ADC = 180° [Linear pair]

⇒ ∠ADB + ∠ADB = 180°

⇒ ∠ADB = ∠ADC = 90°

⇒ AD ⊥ BC

Since, AD ⊥ BC, and BD = BC

∴ AD is perpendicular bisector of BC and as well as altitude from A to BC.

Now,

∠BAD = ∠CAD [By CPCT]

AD is angle bisector of ∠A.

∴ AD is median, perpendicular bisector, altitude and angle- bisector.

Answer 2

Answer:

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