Please solve question no.10. (Please don't send me any links....) Share with your friends Share 0 Akshita Arora answered this Dear student Given: ABCD is a parallelogram or ∥gm, and diagonal AC bisects ∠A.(a) As shown in above figure, ∠1=∠2 (given) ..........1Since, AB∥CD and AC is a transversal, then ∠1=∠4 (alternate interior angles) .......2 Since AD∥BC and AC is a transversal, thenand again, ∠2=∠3 (alternate interior angles) .........3then, from the equation (1),∠1=∠2∠4=∠3 Using 2 and 3 ∴ AC bisect ∠C(b) We have,∠1 = ∠2 givenAlso, from 3, we have ∠2=∠3Hence, we have∠1 = ∠3In ∆ABC, ∠1 = ∠3⇒BC = AB sides opposite to equal sides are equal ........4In ∆AOB and ∆COB,AB = BC Proved aboveOB = OB CommonAO = OC Diagonals of ∥gm bisect each other∆AOB ≅ ∆COBSSS⇒∠AOB = ∠BOC CPCTNow, ∠AOB + ∠BOC = 180° Linear pair⇒∠AOB = ∠BOC = 90°⇒AC⊥BD.(c).Since, ABCD is a parallelogram, soAB = CD and AC = BD .......5Also, AB = BC Using 4 ........6Now, from 5 and 6, we getAB = BC = CD = AD.So, ABCD is a parallelogram in which adjacent sides are equal and also diagonals are ⊥ to each other.So, ABCD is a rhombus. Regards 0 View Full Answer
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