Question

Diagonal AC of a parallelogram ABCD bisects $\angle \text{A}$. Show that
[1] It bisects $\angle \text{C}$ also.
[2] ABCD is a rhombus.

Answer 1

Hint: Use the property that the opposite sides of a parallelogram are parallel. Hence prove that the angles DAC and ACB are equal. Similarly, prove that the angles BAC and DCA are equal. Hence prove that AC bisects angle C also.

Given: ABCD is a parallelogram and AC bisects $\angle \text{A}$
To prove: AC bisects $\angle \text{C}$ .
Proof:
Since ABCD is a parallelogram, we have AB||CD and AD||BC.
Since AD||BC, we have $\angle \text{DAC=}\angle \text{ACB}$ (Alternate interior angles) (i)
Since AB||DC, we have $\angle \text{BAC=}\angle \text{DCA}$ (Alternate interior angles) (ii)
Now since AC bisects $\angle \text{A}$, we have
$\angle \text{DAC=}\angle \text{BAC (iii)}$
From equations (i),(ii) and (iii), we have
$\angle \text{ACB=}\angle \text{DCA}$
Hence AC bisects $\angle \text{C}$ .
[2] The parallelogram satisfying such property is a rhombus.
Proof:
We know that $\angle \text{DAC=}\angle \text{BAC}$(from equation iii) and $\angle \text{BAC=}\angle \text{DCA}$ (from equation ii)
Hence we have $\angle \text{DAC=}\angle \text{DCA}$
Hence in triangle ACD, we have $\angle \text{DAC=}\angle \text{DCA}$.
Hence triangle ACD is an isosceles triangle and hence AD =DC. (Equal angles have equal opposite sides).
But AD=BC and DC = AB
Hence AB = AD = BC = CD.
Hence ABCD is a rhombus.

Note: [1] Alternatively, we know that a diagonal of a parallelogram divides the parallelogram into two triangles of congruent area.
Hence, we have$\Delta \text{ACB}\cong \Delta \text{CAD}$
Hence $\angle \text{DAC=}\angle \text{ACB}$ and $\angle \text{BAC=}\angle \text{DCA}$ [Corresponding parts of congruent triangles]
Now since AC bisects $\angle \text{A}$, we have
$\angle \text{DAC=}\angle \text{BAC}$
Hence $\angle \text{ACB=}\angle \text{DCA}$
Hence AC bisects $\angle \text{C}$ .
Now consider triangles ABC and ADC.
We have
$\angle \text{ACB=}\angle \text{DCA}$ (because AC bisects $\angle \text{C}$ )
AC = AC (common side)
$\angle \text{DAC=}\angle \text{BAC}$ (because AC bisects $\angle \text{A}$)
Hence $\Delta \text{ABC}\cong \Delta \text{ADC}$ by ASA congruence criterion.
Hence AB = AD .
Hence we have AB = AD = BC = DC.
Hence ABCD is a rhombus.