Question

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​

Answer 1

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.

I hope it helps