Abcd is a trapezium with ab parallel to dc

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• • • • Question:4 D (2,6) C(8,3) X B(5,-3) The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is 90°. The coordinates of A, B and Care (2, 6), (5. -3) and (8, 3) respectively. (i) Find the equation of AD. [3] (ii) Find, by calculation, the coordinates of D. [3] The point E is such that ABCE is a parallelogram. (iii) Find the length of BE. [2] 4 D (2,6) C(8,3) X B(5,-3) The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is 90°. The coordinates of A, B and Care (2, 6), (5. -3) and (8, 3) respectively. (i) Find the equation of AD. [3] (ii) Find, by calculation, the coordinates of D. [3] The point E is such that ABCE is a parallelogram. (iii) Find the length of BE. [2] Previous question Next question

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ∆ABC ≅ ∆BAD (iv) diagonal AC = diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] Solution: Let us join BD and AC in the figure tas shown below.. ADCE is a (i) AD = CE (Opposite sides of parallelogram AECD are equal) However, AD = BC (Given) Therefore, BC = CE ∠CEB = ∠CBE (Angles opposite to equal sides in a triangle are also equal) Consider, ∠BAD + ∠CEB = 180° [Co-Interior angles] ∠BAD + ∠CBE = 180°... (1) [Since,∠CEB = ∠CBE] However, ∠ABC + ∠CBE = 180° ( From Equations (1) and (2), we see that ∠BAD = ∠ABC Thus, ∠A= ∠B (ii) AB || CD ∠A + ∠D = 180° (Angles on the same side of the Also, ∠C + ∠B = 180° (Angles on the same side of the transversal) ∴∠A + ∠D = ∠C + ∠B However, ∠A = ∠B [Using the result obtained in (i)] ∴∠C = ∠D (iii) In ∆ABC and ∆BAD, AB = BA (Common side) BC = AD (Given) ∠B = ∠A (Proved before) ∴ ∆ABC ≅ ∆BAD ( (iv) Since∆ABC ≅ ∆BAD, ∴ AC = BD (By CPCT) ☛ Check: Video Solution: ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ∆ABC ≅ ∆BAD (iv) diagonal AC = diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] Summary: If ABCDis a trapezium in which AB || CD and AD = BC, we have proved that ∠A = ∠B, ∠C = ∠D,△ABC≅△BAD by SAS congruence, anddiagonal AC =diagonal BD. ☛ Related Questions: • • • •

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ABCD is a trapezium in which AB || DC, BD is a diagonaland E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. Solution: Let EF intersect DB at G as shown below. By converse of In trapezium ABCD, EF || AB and E is the mid-point of AD. Therefore, G is the mid-point of DB. [Converse of mid-point theorem] As EF || AB and AB || CD, ∴ EF || CD (Two lines parallel to the same line are In ΔBCD, GF || CD and G is the mid-point of line BD. Therefore, by using the converse of mid-point theorem, F is the Video Solution: ABCD is a trapezium in which AB || DC, BD is a diagonaland E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC Summary: If ABCDis a trapezium in which AB || CD, BD is a diagonal and E is mid-point of AD, a line is drawn through Eparallel to AB intersecting BC at F, then F is the mid-point of BC. ☛ Related Questions: • • • •