ABCD is a trapezium in which AB is parallel to DC and AD =BC. If CE is

1.	ABCD is a trapezium in which AB is parallel to DC and AD =BC. If CE is drawn parallel to AD meeting AB at E prove the following: a) AECD is a parallelogram b) AD=EC c) Triangle CEB is an isosceles triangle( Please explain in detail and don't answer if u don't know otherwise ur answer will be reported)
Answer» Answer: Let US extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i) AD = CE (OPPOSITE sides of parallelogram AECD) However, AD = BC (Given) Therefore, BC = CE ∠CEB = ∠CBE (Angle opposite to equal sides are also equal) Consider parallel lines AD and CE. AE is the transversal line for them. ∠A + ∠CEB = 180º (Angles on the same side of transversal) ∠A + ∠CBE = 180º (Using the relation∠CEB = ∠CBE) ... (1) However, ∠B + ∠CBE = 180º (Linear pair angles) ... (2) From equations (1) and (2), we obtain ∠A = ∠B (ii) AB \|\| CD ∠A + ∠D = 180º (Angles on the same side of the transversal) 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 ΔABC and ΔBAD, AB = BA (Common side) BC = AD (Given) ∠B = ∠A (Proved before) ∴ ΔABC ≅ ΔBAD (SAS CONGRUENCE rule) (iv) We had observed that, ΔABC ≅ ΔBAD ∴ AC = BD (By CPCT)

ABCD is a trapezium in which AB is parallel to DC and AD =BC. If CE is drawn parallel to AD meeting AB at E prove the following: a) AECD is a parallelogram b) AD=EC c) Triangle CEB is an isosceles triangle( Please explain in detail and don't answer if u don't know otherwise ur answer will be reported)

Answer»

Answer:

Let US extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram.

(i) AD = CE (OPPOSITE sides of parallelogram AECD)

However, AD = BC (Given)

Therefore, BC = CE

∠CEB = ∠CBE (Angle opposite to equal sides are also equal)

Consider parallel lines AD and CE. AE is the transversal line for them.

∠A + ∠CEB = 180º (Angles on the same side of transversal)

∠A + ∠CBE = 180º (Using the relation∠CEB = ∠CBE) ... (1)

However, ∠B + ∠CBE = 180º (Linear pair angles) ... (2)

From equations (1) and (2), we obtain

∠A = ∠B

(ii) AB || CD

∠A + ∠D = 180º (Angles on the same side of the transversal)

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 ΔABC and ΔBAD,

AB = BA (Common side)

BC = AD (Given)

∠B = ∠A (Proved before)

∴ ΔABC ≅ ΔBAD (SAS CONGRUENCE rule)

(iv) We had observed that,

ΔABC ≅ ΔBAD

∴ AC = BD (By CPCT)