Prove the mid point theorem and the inverse of that theorem.
Answers
∠QAP = ∠QCR. (Pair of alternate angles) ---------- (1)
AQ = QC. (∵ Q is the mid-point of side AC) ---------- (2)
∠AQP = ∠CQR (Vertically opposite angles) ---------- (3)
Thus, ΔAPQ ≅ ΔCRQ (ASA Congruence rule)
PQ = QR. (by CPCT). or PQ = 1/ 2 PR ---------- (4)
⇒ AP = CR (by CPCT) ........(5)
But, AP = BP. (∵ P is the mid-point of the side AB)
⇒ BP = CR
Also. BP || CR. (by construction)
In quadrilateral BCRP, BP = CR and BP || CR
Therefore, quadrilateral BCRP is a parallelogram.
BC || PR or, BC || PQ
Also, PR = BC (∵ BCRP is a parallelogram)
⇒ 1 /2 PR = 1/ 2 BC
⇒ PQ = 1/ 2 BC. [from (4)]
hi mate,
Construction- Extend the line segment DE and produce it to F such that, EF=DE.
In the triangle, ADE, and also the triangle CFE
EC= AE —– (given)
∠CEF = ∠AED {vertically opposite angles}
EF = DE { by construction}
hence,
△ CFE ≅ △ ADE {by SAS}
Therefore,
∠CFE = ∠ADE {by c.p.c.t.}
∠FCE= ∠DAE {by c.p.c.t.}
and CF = AD {by c.p.c.t.}
The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.
In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.
Therefore, CF ∥ AB
So, CF ∥ BD
and CF = BD {since BD = AD, it is proved that CF = AD}
Thus, BDFC forms a parallelogram.
By the use of properties of a parallelogram, we can write
BC ∥ DF
and BC = DF
BC ∥ DE
and DE = (1/2 * BC).
Hence, the midpoint theorem is Proved.
The converse of MidPoint Theorem
The converse of the midpoint theorem states that ” if a line is drawn through the midpoint of one side of a triangle, and parallel to the other side, it bisects the third side”.