Question

Explain Mid point theorem.

Answer 1

Mid-Point Theorem

The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

In triangle ABC, P and Q are mid-points of AB and AC respectively.
We want to Prove that
i) $PQ\parallel BC$
ii) $PQ=\left(\frac{1}{2}\right)BC$
Let us draw $CR\parallel BA$ to meet PQ produced at R.

$\angle QAP=\angle QCR$ (Pair of alternate angles) ...... (1)
AQ = QC ($\because$ Q is the mid point of side AC) ...... (2)
$\angle AQP=\angle CQR$ (Vertically opposite angles) ....... (3)
Thus, $\Delta APQ\cong \Delta CRQ$ (ASA Congruence rule)
PQ = QR (by CPCT) or $PQ=\left(\frac{1}{2}\right)PR$ ..... (4)
$\Rightarrow$ AP = CR (by CPCT) ..... (5)
But, AP = BP ($\because$ P is the mid-point of the side AB)
$\Rightarrow$ BP = CR
Also, $BP\parallel CR$ (by construction)
In quadrilateral BCRP, BP = CR and $BP\parallel CR$
Therefore, quadrilateral BCRP is a parallelogram
$BC\parallel PR$ or $BC\parallel PQ$
Also, PR = BC ($\because$ BCRP is a parallelogram)
$\Rightarrow \left(\frac{1}{2}\right)PR=\left(\frac{1}{2}BC\right)$
$\Rightarrow PQ=\left(\frac{1}{2}\right)BC$ [from (4)]