Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th
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(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A. ∴  ∠DAC=∠BAC      ---- ( 1 ) Now, AB∥DC and AC as traversal, ∴  ∠BAC=∠DCA          [ Alternate angles ]  --- ( 2 ) AD∥BC and AAC as traversal, ∴  ∠DAC=∠BCA         [ Alternate angles ]   --- ( 3 ) From ( 1 ), ( 2 ) and ( 3 ) ∠DAC=∠BAC=∠DCA=∠BCA ∴  ∠DCA=∠BCA Hence, AC bisects ∠C. (ii)  In △ABC, ⇒  ∠BAC=∠BCA      [ Proved in above ] ⇒  BC=AB      [ Sides opposite to equal angles are equal ]    --- ( 1 ) ⇒  Also, AB=CD and AD=BC     [ Opposite sides of parallelogram are equal ]       ---- ( 2 ) From ( 1 ) and ( 2 ), ⇒  AB=BC=CD=DA Hence, ABCD is a rhombus.
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In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

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Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Description : The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Last Answer : According to question parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°.