Question

Given: BD bisects AC and ∠CBE≅∠ADE.

Answer 1

Here

• BC=AD(Parallel)
• <BEC=<AEC
• <CBE=<ADE

So

∆BEC$\cong$∆AED(ASA)

Also similarly

• ∆ABE$\cong$∆CBE

Hence it's a paralleogram

Answer 2

Answer:

know that,

Sum of all four angles of a quadrilateral = 360o

Sum of two given angles = 40o + 110o = 150o

So, the sum of remaining two angles = 360o – 150o = 210o

Also given,

Ratio in these angles = 3 : 4

Hence,

Third angle = (210o x 3)/(3 + 4)

= (210o x 3)/7

= 90o

And,

Fourth angle = (210o x 4)/(3 + 4)

= (210o x 4)/7