ABCD is a parallelogram. If the diagonal AC bisects ZA then prove that
(a) AC will bisect C
(b) AC I BD
(c) ABCD is a rhombus
Answers
Answer:
(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A.
(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A.∴ ∠DAC=∠BAC ---- ( 1 )
(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A.∴ ∠DAC=∠BAC ---- ( 1 )Now,
(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A.∴ ∠DAC=∠BAC ---- ( 1 )Now,AB∥DC and AC as traversal,
(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 )
(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,
(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 )
(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 )
(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
(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
(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=∠BCAHence, AC bisects ∠C.