Question

In the adjoining figure, ABCD and PQRC are rectangles, where Q is the mid-point of AC. Prove that (i) DP = PC (ii) $PR=\frac{1}{2}AC$.

Answer 1

(i)∠CRQ = ∠CBA = 90^o ⇒ QR ∣∣​ AB
In ∆ABC, Q is the mid point of AC and QR ∣∣​ AB​.
∴​ R is the mid point of BC.
Similarly, P is the midpoint of DC.
∴​ DP = PC

(ii) Join BD. In ∆​CDB, P is the mid point of DC and R is the mid point of BC.
∴​ PR ∣∣​ DB and PR = $\frac{1}{2}$DB = $\frac{1}{2}$AC [∵​ AC = BD]