The side AC of a triangle ABC is produced to point E so that CE = . D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that : (i) 3DF = EF (ii) 4CR = AB.

Question

KnowledgeBoat · Accepted Answer

By mid-point theorem, The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it. By converse of mid-point theorem, The straight line drawn through the mid-point of one side of a triangle parallel to another, bisects the third side. By equal intercept theorem, If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts. (i) In △ ABC, D is the mid-point of BC and DP || AB. ∴ P is the mid-point of AC. (By converse of mid-point theorem) ∴ AP = PC Given, ⇒ CE = ∴ CE = PC Since, AP = PC and CE = PC, ∴ AP = PC = CE. Since, ⇒ AB || DP || CR ⇒ AF || DP || CR (Since, point F lies on straight line AB) In △ AEF, AF || PD || CR and AP = PC = CE ∴ DF = DR = RE = x (let) [By equal intercept theorem] From figure, ⇒ EF = DF + DR + RE = x + x + x = 3x = 3DF. Hence, proved that 3DF = EF. (ii) In △ ABC, D is mid-point of BC and DP || AB. ∴ P is the mid-point of AC. (By converse of mid-point theorem) ∴ PD = (By mid-point theorem) .........(1) In △ PED, Given, ⇒ CE = ⇒ CE = PC (Since, P is mid-point of AC) ∴ C is the mid-point of PE. C is mid-point of PE and DP || CR. ∴ R is the mid-point of DE. (By converse of mid-point theorem) ∴ CR = (By mid-point theorem) .........(2) Substituting value of PD from equation (1) in equation (2), we get : ⇒ CR = ⇒ CR = ⇒ 4CR = AB. Hence, proved that 4CR = AB.

Mathematics

The side AC of a triangle ABC is produced to point E so that CE = . D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that : (i) 3DF = EF (ii) 4CR = AB.

Mid-point Theorem

Related Questions

Use the following figure to find :
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm.
(iv) DF, if CG = 11 cm.

In the figure, given below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that :
(i) AQ // BS
(ii) DS = 3RS

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :
(i) M, A and N are collinear.
(ii) A is the mid-point of MN.

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.

Mathematics

The side AC of a triangle ABC is produced to point E so that CE = . D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that : (i) 3DF = EF (ii) 4CR = AB.

Mid-point Theorem

Related Questions

Use the following figure to find :(i) BC, if AB = 7.2 cm.(ii) GE, if FE = 4 cm.(iii) AE, if BD = 4.1 cm.(iv) DF, if CG = 11 cm.

In the figure, given below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that :(i) AQ // BS(ii) DS = 3RS

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :(i) M, A and N are collinear.(ii) A is the mid-point of MN.

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.

Use the following figure to find :
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm.
(iv) DF, if CG = 11 cm.

In the figure, given below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that :
(i) AQ // BS
(ii) DS = 3RS

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :
(i) M, A and N are collinear.
(ii) A is the mid-point of MN.