Question

in triangle ABC p, q, r are the midpoint of sides AB, BC, CA respectively. Prove that the straight lines PQ, QR, RP form four congruent triangle

Answer 1

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In Δ ABC

P is the midpoint of AB

R is the midpoint of CA

∴ By midpoint theorem,

PR ║ BC

PR = $\frac{1}{2}$ BC

Since Q is the midpoint of BC,

PR = BC

∴ PBQR is a Parallelogram

(One pair of opposite sides are equal (PR = BC) and parallel (PR║BC)

Similarly,

PRCQ is a parallelogram

APQR is a parallelogram

Now,

In Parallelogram PBQR

ΔPBQ ≅ ΔPRQ → 1

(Diagonals of a parallelogram divide it into two congruent triangles)

In Parallelogram PRCQ

ΔPRQ ≅ ΔCRQ → 2

(Diagonals of a parallelogram divide it into two congruent triangles)

In Parallelogram APQR

ΔAPR ≅ ΔQPR → 3

(Diagonals of a parallelogram divide it into two congruent triangles)

Now,

From 1, 2 and 3

ΔPBQ ≅ ΔPRQ ≅ ΔCQR ≅ ΔAPR

Therefore, the linesegments joining the midpoints of the sides of a triangle, form four congruent triangles.

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