ab and CD are respectively the smallest and longest sides of a quadrilateral ABCD show that angle A is greater than angle C and Angle B is greater than angle d
Attachments:
kvnmurty:
Report an answer when they write improper contents.. or incorrect conents... @yg8...
Answers
Answered by
23
Join AC.
In triangle ABC, AB < BC. So ∠ACB < ∠BAC.
In triangle ADC, AD < CD. So ∠ACD < ∠DAC.
Add above inequalities, ∠ACB + ∠ACD < ∠BAC + ∠DAC
So ∠C < ∠A Proved.
Similarly now draw the diagonal BD. Follow above procedure, by splitting angles at B and D into two parts.
In triangle ABC, AB < BC. So ∠ACB < ∠BAC.
In triangle ADC, AD < CD. So ∠ACD < ∠DAC.
Add above inequalities, ∠ACB + ∠ACD < ∠BAC + ∠DAC
So ∠C < ∠A Proved.
Similarly now draw the diagonal BD. Follow above procedure, by splitting angles at B and D into two parts.
:-)
Similar questions