In ΔPQR, if PT is the median, then which of the following is correct?

Correct — B. PT is the median, so T is the midpoint of QR and QT = TR. That is exactly the condition Apollonius' theorem is stated under.

Apollonius: PQ² + PR² = 2PT² + 2QT²
which is option (b), PQ² + PR² = 2(PT² + QT²).

Test it on an equilateral triangle of side 2, where PT = √3 and QT = 1:
LHS = 2² + 2² = 8
RHS = 2(3 + 1) = 8

Why the others are wrong

• (a)
  PQ² + PR² = PT² + QR² loses the factor 2 and uses the whole side QR where the theorem uses half of it. The equilateral test gives 3 + 4 = 7 against a true 8.
• (c)
  PQ² + PR² = PT² + QT² is the right pair of terms with the 2 stripped off. The equilateral test gives 3 + 1 = 4, half of the correct 8.
• (d)
  PQ² + PR² = 2(PT² − QT²) flips the sign inside the bracket. Subtracting QT² gives 2(3 − 1) = 4 on the equilateral test, again half of 8.

Apollonius' theorem ties the two sides of a triangle to the median drawn to the third: PQ² + PR² = 2(PT² + QT²), where T is the midpoint of QR so that QT = QR⁄2.

Putting QT = QR⁄2 into it gives the other standard form, PQ² + PR² = 2PT² + QR²⁄2 — the same statement, easier to use when a question hands you the whole third side.

It follows from the cosine rule on triangles PQT and PRT. The angles at T are supplementary, so their cosine terms are equal and opposite and cancel the moment you add the two equations.

Key facts

• Apollonius' theorem: PQ² + PR² = 2(PT² + QT²) when PT is the median to QR.
• T is a midpoint, so QT = TR = QR⁄2, giving the equivalent form PQ² + PR² = 2PT² + QR²⁄2.
• In an equilateral triangle of side 2 the median is √3, and both sides of the theorem come to 8.
• The theorem is the cosine rule applied to the two halves, where the supplementary angles at the foot cancel.

Common traps

• Writing QR where the theorem needs QT, which is half of it.
• Dropping the factor 2 outside the bracket.
• Assuming the median is also an altitude, which is true only when PQ = PR.

No directly related past PYQ was found.