Question

6. Diagonal AC of a parallelogram ABCD bisects
ZA (see Fig. 8.19). Show that
(i) it bisects 2 C also,
(ii) ABCD is a rhombus.​

Answer 1

Given:

• a parrallogram ABCD

• diagonal AC bisects ∠A

To Prove:

• it bisects ∠C

• ABCD is a rhombus

Solution:

1) here,ABCD is a parrallogram and AC bisect ∠A

$∴$∠DAC = ∠BAC ...( 1 )

Now,

AB||DC and AC as transversal,

$∴$∠BAC = ∠DAC ...(2) [alternated angles]

AB||BC and AAC as transversal,

$∴$∠DAC = ∠BCA ...(3) [alternated angles]

From ( 1 ),(2) and (3)

∠DAC = ∠BAC = ∠DCA = ∠BCA

⊱ $\tt\underline\red{hence,\:AC\: bisects\:∠C}$

2) in ∆ABC,

$⟹$∠BAC = ∠BCA [proved in above]

$⟹$ BC = AB ..... ( 1 )[side opposite to equal angles are equal]

also,

$⟹$AB = CD and AD = BC .....( 2)[opposite sides of parrallogram are equal]

From ( 1 ) and (2),

$⟹$AB = BC = CD = DA

⊱ $\tt\underline\red{hence,\:ABCD\:is\:a\: rhombus}$

Hence proved ✓

Answer 2

AB = BC = CD = DA

 

Hence proved ✓