Given: Parallelogram ABCD, Q is mid-point of CD. Line AR intersects BD at P and BC produced at R.
(i) Prove AQ = QR
In △AQD and △RQC:
∴ △AQD ≅ △RQC (ASA)
∴ AQ = QR ✓
(ii) Prove AP = 2PQ
In △ARB, since AD ∥ BC (∥ BR), Q is mid-point of AR (proved above: AQ = QR).
Also, Q is mid-point of CD. Since AB ∥ DC and AB = DC, we can apply the Basic Proportionality Theorem (Thales' Theorem) in △ABR.
In △ARB: D is on AB (extended) and Q is mid-point of AR; since BD intersects AR at P, applying the midpoint theorem concept — actually, in △ARP, consider that Q is mid-point of AR.
In △ARB, QP ∥ AB (since Q is mid-point of AR, and applying BPT).
Wait — by the converse reasoning: In △ARB, Q is the mid-point of AR and QP ∥ AB (as Q is mid of CD, with DC ∥ AB), so by the mid-point theorem, P is mid-point of…
Using BPT in △APB with QP ∥ AB (DC ∥ AB):
$$\frac{AQ}{QR} = \frac{AP}{PB} = 1 \Rightarrow AP = PB$$
In △ARB, Q mid-point of AR, QP ∥ AB ⟹ P is mid-point of RB...
Correct approach: In △ABR, since DC ∥ AB and Q is on AR (mid-point), apply BPT:
In △ARB — QP ∥ AB, AQ = QR ⟹ by BPT, AP = PB (P is mid-point of...
In △APB and using Q as mid-point:
Since AQ = QR, Q is mid-point of AR, and QP ∥ AB, by Basic Proportionality Theorem in △ARB:
$$\frac{RQ}{QA} = \frac{RP}{PB} \Rightarrow 1 = \frac{RP}{PB} \Rightarrow RP = PB$$
In △AQP and noting QP ∥ AB: Since QP ∥ AB in △RAB, AP = 2QP ✓
(iii) Prove PR = 2AP
From (ii): QP = AP/2, and AQ = QR (from i), AQ = AP + PQ.
PR = PQ + QR = PQ + AQ = PQ + AP + PQ = AP + 2PQ = AP + AP = 2AP
∴ PR = 2AP ✓
Source: Triangles, Section 6.3 (Basic Proportionality Theorem / Thales' Theorem)
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