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A triangle PQR is formed by joining the mid-points of the sides of a triangle ABC. 'O' is the circumcentre of
ΔABC, then for ΔPQR, the point 'O' is :
- a)incentre
- b)circumcentre
- c)orthocentre
- d)centroid
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A triangle PQR is formed by joining the mid-points of the sides of a t...
Triangle ABC. Prove that triangle PQR is congruent to triangle ABC.
Proof:
Let M, N, and O be the midpoints of sides AB, BC, and AC respectively. Then, by definition, we have:
PM = MQ = MA/2
QN = NR = NB/2
PR = RP = OC/2
Since O is the circumcenter of triangle ABC, it is equidistant from all three vertices, i.e.
OA = OB = OC
Using this fact, we can rewrite the above expressions as:
PM = MQ = (OA + OB)/2
QN = NR = (OB + OC)/2
PR = RP = (OC + OA)/2
Adding all three equations, we get:
PM + QN + PR = (OA + OB + OB + OC + OC + OA)/2
= (2OA + 2OB + 2OC)/2
= OA + OB + OC
But OA + OB + OC is just the perimeter of triangle ABC, so we can rewrite the above equation as:
PQ + QR + RP = AB + BC + AC
This means that triangle PQR has the same perimeter as triangle ABC. Furthermore, we note that the three sides of triangle PQR are parallel to the corresponding sides of triangle ABC, and are half their length. Therefore, triangle PQR is similar to triangle ABC, with a scale factor of 1/2.
To show that the two triangles are congruent, we need to prove that they have the same angles. Since the sides of triangle PQR are parallel to the corresponding sides of triangle ABC, it follows that the corresponding angles are equal. Therefore, triangle PQR is congruent to triangle ABC.
Proof:
Let M, N, and O be the midpoints of sides AB, BC, and AC respectively. Then, by definition, we have:
PM = MQ = MA/2
QN = NR = NB/2
PR = RP = OC/2
Since O is the circumcenter of triangle ABC, it is equidistant from all three vertices, i.e.
OA = OB = OC
Using this fact, we can rewrite the above expressions as:
PM = MQ = (OA + OB)/2
QN = NR = (OB + OC)/2
PR = RP = (OC + OA)/2
Adding all three equations, we get:
PM + QN + PR = (OA + OB + OB + OC + OC + OA)/2
= (2OA + 2OB + 2OC)/2
= OA + OB + OC
But OA + OB + OC is just the perimeter of triangle ABC, so we can rewrite the above equation as:
PQ + QR + RP = AB + BC + AC
This means that triangle PQR has the same perimeter as triangle ABC. Furthermore, we note that the three sides of triangle PQR are parallel to the corresponding sides of triangle ABC, and are half their length. Therefore, triangle PQR is similar to triangle ABC, with a scale factor of 1/2.
To show that the two triangles are congruent, we need to prove that they have the same angles. Since the sides of triangle PQR are parallel to the corresponding sides of triangle ABC, it follows that the corresponding angles are equal. Therefore, triangle PQR is congruent to triangle ABC.
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A triangle PQR is formed by joining the mid-points of the sides of a t...
C
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A triangle PQR is formed by joining the mid-points of the sides of a triangle ABC. 'O' is the circumcentre ofΔABC, then for ΔPQR, the point 'O' is :a)incentreb)circumcentrec)orthocentred)centroidCorrect answer is option 'C'. Can you explain this answer?
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