Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.12). Show that: 

(i) ∆APD ≅ ∆CQB   

(ii) AP = CQ 

(iii) ∆AQB ≅∆CPD 

(iv) AQ = CP 

(v) APCQ is a parallelogram

 

Answer 1

Consider the following geometric situation:

We are given a parallelogram \( ABCD \) with the following properties:

Step (i): Congruence of Triangles \( \Delta APD \) and \( \Delta CQB \)

In triangles \( \Delta APD \) and \( \Delta CQB \), the following properties hold:

\[ \angle ADP = \angle CBQ \quad \text{(Alternate interior angles for \( BC \parallel AD \))} \] \[ AD = CB \quad \text{(Opposite sides of parallelogram \( ABCD \))} \] \[ DP = BQ \quad \text{(Given)} \] \[ \therefore \Delta APD \cong \Delta CQB \quad \text{(By SAS congruence rule)} \]

Step (ii): Conclusion from Congruent Triangles

Since \( \Delta APD \cong \Delta CQB \), we can conclude the following:

\[ AP = CQ \quad \text{(By CPCT, Corresponding Parts of Congruent Triangles)} \]

Step (iii): Congruence of Triangles \( \Delta AQB \) and \( \Delta CPD \)

Now, consider the triangles \( \Delta AQB \) and \( \Delta CPD \), where:

\[ \angle ABQ = \angle CDP \quad \text{(Alternate interior angles for \( AB \parallel CD \))} \] \[ AB = CD \quad \text{(Opposite sides of parallelogram \( ABCD \))} \] \[ BQ = DP \quad \text{(Given)} \] \[ \therefore \Delta AQB \cong \Delta CPD \quad \text{(By SAS congruence rule)} \]

Step (iv): Conclusion from Congruent Triangles

Since \( \Delta AQB \cong \Delta CPD \), we can conclude:

\[ AQ = CP \quad \text{(By CPCT, Corresponding Parts of Congruent Triangles)} \]

Step (v): Conclusion for Parallelogram

From the results obtained in steps (ii) and (iv), we have:

\[ AQ = CP \quad \text{and} \quad AP = CQ \]

Since opposite sides in quadrilateral \( APCQ \) are equal to each other, it follows that \( APCQ \) is a parallelogram.