Question

a quadrilateral ABCD is designed such that angle 1 equal Angle 3 and angle 2 equal angle 4. if AE/EC=DE/BC=1/2 and DC = 4 cm then AB equal

please tell me it's explanation.. it's a humble request​

Answer 1

Answer:

Given : ABCD is A parallelogram

To Prove : BF=BC

Proof : In △DCE,DE=DC (given)

∴∠DCE=∠DEC...(1)

(Equal sides have equal is opposite to them)

since,

AB∥CD,∠DCE=∠BFC...(2) (pair of corresponding ∠S)

Form (1) and (2)

∠DEC=∠BFC

In △AEF,∠AEF=∠AFE

∴AF=AE,

⇒AB+BF=AD+DE

⇒BF=AD [∵AB=CD=DE]

⇒BF=BC [∵AD=BC] Hence proved.

Answer 2

Answer:

8cm

Explanation:

In triangle DEC and AEB

angle 1 = angle 3

angle 2 = angle 4

by AA similarity triangle DEC is similar to triangle AEB

===> AE/EC = BE/DE = AB/CD = 2/1

AB/CD = 2/1

AB/4 = 2/1 ( CD = 4cm given )

AB = 8cm

Think so it will be help full for your studies

All the best