In the adjoining figure, ABCD is a parallelogram, E is the mid-point of CD and through D, a line is drawn parallel to EB to meet CB produced at G and intersecting AB at F. Prove that : (i) AD = GC (ii) DG = 2 EB

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 side of a triangle parallel to another, bisects the third side. Given, DF || EB ⇒ DG || EB In △ DGC, ⇒ E is the mid-point of CD and DG || EB. ∴ B is the mid-point of GC. (By converse of mid-point theorem) ∴ BG = BC Thus, BC = (i) We know that, Opposite sides of a parallelogram are equal. ⇒ AD = BC ⇒ AD = GC Hence, proved that AD = GC. (ii) In △DGC, Since, E and B are the mid-points of DC and GC respectively. EB || DG Thus, by mid-point theorem, ⇒ EB = DG ⇒ DG = 2 EB. Hence, proved that DG = 2EB.

Mathematics

In the adjoining figure, ABCD is a parallelogram, E is the mid-point of CD and through D, a line is drawn parallel to EB to meet CB produced at G and intersecting AB at F. Prove that : (i) AD = GC (ii) DG = 2 EB

Mid-point Theorem

Related Questions

In the adjoining figure, ABCD is a kite in which AB = AD and CB = CD. If E, F, G are respectively the mid-points of AB, AD and CD, prove that :
(i) ∠EFG = 90°
(ii) If GH || FE, then H bisects CB.

In the adjoining figure, △ABC is right-angled at B and P is the mid-point of AC. Show that, PA = PB = PC.

Mathematics

In the adjoining figure, ABCD is a parallelogram, E is the mid-point of CD and through D, a line is drawn parallel to EB to meet CB produced at G and intersecting AB at F. Prove that : (i) AD = GC (ii) DG = 2 EB

Mid-point Theorem

Related Questions

In the adjoining figure, ABCD is a kite in which AB = AD and CB = CD. If E, F, G are respectively the mid-points of AB, AD and CD, prove that :(i) ∠EFG = 90°(ii) If GH || FE, then H bisects CB.

In the adjoining figure, △ABC is right-angled at B and P is the mid-point of AC. Show that, PA = PB = PC.

In the adjoining figure, ABCD is a kite in which AB = AD and CB = CD. If E, F, G are respectively the mid-points of AB, AD and CD, prove that :
(i) ∠EFG = 90°
(ii) If GH || FE, then H bisects CB.