Question

using theorem 6.1 prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisect the third side​??

Answer 1

DE || BC
DE = 1/2 BC
Construction: Draw CR || BA to meet DE produced at R. (Refer the above figure)

∠EAD = ∠ECR. (Pair of alternate angles) ———- (1)
AE = EC. (∵ E is the mid-point of side AC) ———- (2)
∠AEP = ∠CQR (Vertically opposite angles) ———- (3)
Thus, ΔADE ≅ ΔCRE (ASA Congruence rule)

DE = 1/2 DR ———- (4)
But, AD= BD. (∵ D is the mid-point of the side AB)
Also. BD || CR. (by construction)

In quadrilateral BCRD, BD = CR and BD || CR
Therefore, quadrilateral BCRD is a parallelogram.
BC || DR or, BC || DE
Also, DR = BC (∵ BCRD is a parallelogram)
⇒ 1/2 DR = 1/2 BC

Answer 2

Answer:

DE || BC

DE = 1/2 BC

Construction: Draw CR || BA to meet DE produced at R. (Refer the above figure)

∠EAD = ∠ECR. (Pair of alternate angles) ———- (1)

AE = EC. (∵ E is the mid-point of side AC) ———- (2)

∠AEP = ∠CQR (Vertically opposite angles) ———- (3)

Thus, ΔADE ≅ ΔCRE (ASA Congruence rule)

DE = 1/2 DR ———- (4)

But, AD= BD. (∵ D is the mid-point of the side AB)

Also. BD || CR. (by construction)

In quadrilateral BCRD, BD = CR and BD || CR

Therefore, quadrilateral BCRD is a parallelogram.

BC || DR or, BC || DE

Also, DR = BC (∵ BCRD is a parallelogram)

⇒ 1/2 DR = 1/2 BC