ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Given: ABCD is a rhombus.
To Prove: (i) Diagonal AC bisects ∠A as well as ∠C.
(ii) Diagonal BD bisects ∠B as well as ∠D.
Proof: ∵ ABCD is a rhombus
∴ AD = CD
∴ ∠DAC = ∠DCA ...(1)
| Angles opposite to equal sides of a triangle are equal
Also, AD || BC
and transversal AC intersects them
∴ ∠DAC = ∠BCA ...(2)
| Alt. Int. ∠s
From (1) and (2)
∠DCA = ∠BCA
⇒ AC bisects ∠C
Similarly AC bisects ∠A.
(ii) Proceeding similarly as in (i) above, we can prove that BD bisects ∠B as well as ∠D.