Question

ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:
(i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD.
$\left[3Marks\right]$

Answer 1

Given: In parallelogram ABCD in which diagonal AC bisects ∠A.

Proof:

(i) As AB || CD, we have [Opposite sides of a || gm]

∠DCA = ∠CAB

Similarly, ∠DAC = ∠DCB

But, ∠CAB = ∠DAC [Since, AC bisects ∠A]

Hence,

∠DCA = ∠ACB and AC bisects ∠C.
$\left[1Mark\right]$

(ii) As AC bisects ∠A and ∠C

And, ∠A = ∠C

Hence, ABCD is a rhombus.
$\left[1Mark\right]$

(iii) Since, AC and BD are the diagonals of a rhombus and

AC and BD bisect each other at right angles

Hence, AC ⊥ BD
$\left[1Mark\right]$