Given: BD bisects AC and ∠CBE≅∠ADE.
2 answers:
Here
BC=AD(Parallel) <BEC=<AEC <CBE=<ADE So
∆BEC ∆AED(ASA)
Also similarly
∆ABE ∆CBE
Hence it's a paralleogram
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
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