Question

Diagonal AC of a parallelogram ABCD bisects ∠ A (see Fig. 8.11). Show that 

(i) it bisects ∠C also, 

(ii) ABCD is a rhombus.

Answer 1

Given:

ABCD is a parallelogram.

To Prove:

We need to prove that AC bisects ∠C and that ABCD is a rhombus.

Solution:

(i) From the properties of a parallelogram, we have the following relations:

∠DAC = ∠BCA 
(Alternate interior angles) ... (1)

∠BAC = ∠DCA 
(Alternate interior angles) ... (2)

However, it is given that AC bisects ∠A. Thus:

∠DAC = ∠BAC ... (3)

From equations (1), (2), and (3), we obtain:

∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4)

Therefore, ∠DAC = ∠BCA, and hence AC bisects ∠C.

Conclusion (ii):

From equation (4), we obtain:

∠DAC = ∠DCA

By the property of triangles, sides opposite to equal angles are equal. Thus:

DA = DC

However, in a parallelogram, opposite sides are equal. Hence:

DA = BC and AB = CD

Therefore, we have:

∠AB = ∠BC = ∠CD = ∠DA

Thus, ABCD is a rhombus, as all four sides are equal.

Conclusion:

We have shown that AC bisects ∠C, and since all sides of ABCD are equal, ABCD is a rhombus.