Question

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.13). Show that 

(i) ∆ APB ≅ ∆ CQD 

(ii) AP = CQ

Answer 1

In the given proof, we are working with two triangles, \( \Delta APB \) and \( \Delta CQD \).

Step (i): Consider the following:

• \( \angle APB = \angle CQD = 90^\circ \quad \text{(Right angles)}\)
• \( AB = CD \quad \text{(Opposite sides of parallelogram ABCD)}\)
• \( \angle ABP = \angle CDQ \quad \text{(Alternate interior angles for } AB \parallel CD\text{)}\)

Using these facts, we can apply the AAS congruency rule (Angle-Angle-Side rule) to prove that the two triangles are congruent:

\[ \Delta APB \cong \Delta CQD \quad \text{(By AAS congruency)} \]

Step (ii): From the result above, since the triangles \( \Delta APB \) and \( \Delta CQD \) are congruent, we can conclude:

\[ AP = CQ \quad \text{(By CPCT, corresponding parts of congruent triangles are equal)} \]

Therefore, we have proved that \( AP = CQ \).