Circum center of a trangle is same as its centroid proof
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Heyy mate ❤✌✌❤
Here's your Answer....
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GIVEN: An equilateral triangle ABC, Medians AP, BQ, &CR. Their point of concurrency is O, which is the centroid of the triangle.
TO PROVE THAT: Centroid O is the circumcentre of the triangle ABC.
If we prove that centroid O is the circumcentre of the triangle, then it automatically becomes the centre of the circumcircle.
PROOF: Since AP is median, so P is mid point of BC.
ie, BP = PC.
AB = AC ( as triangle ABC is equilateral)
AP=AP ( common side)
Hence triangle ABP is congruent to ACP( by SSS congruence criterion)
=> angle APB = angle angleAPC ( corresponding parts of congruent triangles)
But their sum = 180°
So, each angle has to be 90°.
That shows that AP is perpendicular bisector of BC.
Similarly, pyove that BQ & CR are perpendicular bisectors of AC & AB respectively.
So now, The point of concurrency ‘O' of these perpendicular bisectors becomes circumcentre of the triangle. ( as circumcentre is the point of concurrency of 3 perpendicular bisectors of the sides of the triangle). And this centre is also the centre of circum circle.
This way centroid O coincides with circumcentre O…
[Hence Proved]
✔✔✔✔✔
Here's your Answer....
⤵️⤵️⤵️⤵️⤵️⤵️⤵️⤵️
GIVEN: An equilateral triangle ABC, Medians AP, BQ, &CR. Their point of concurrency is O, which is the centroid of the triangle.
TO PROVE THAT: Centroid O is the circumcentre of the triangle ABC.
If we prove that centroid O is the circumcentre of the triangle, then it automatically becomes the centre of the circumcircle.
PROOF: Since AP is median, so P is mid point of BC.
ie, BP = PC.
AB = AC ( as triangle ABC is equilateral)
AP=AP ( common side)
Hence triangle ABP is congruent to ACP( by SSS congruence criterion)
=> angle APB = angle angleAPC ( corresponding parts of congruent triangles)
But their sum = 180°
So, each angle has to be 90°.
That shows that AP is perpendicular bisector of BC.
Similarly, pyove that BQ & CR are perpendicular bisectors of AC & AB respectively.
So now, The point of concurrency ‘O' of these perpendicular bisectors becomes circumcentre of the triangle. ( as circumcentre is the point of concurrency of 3 perpendicular bisectors of the sides of the triangle). And this centre is also the centre of circum circle.
This way centroid O coincides with circumcentre O…
[Hence Proved]
✔✔✔✔✔
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