# Selina Solutions Class 9 Mathematics Solutions for Exercise 13(A) in Chapter 13 - Chapter 13- Pythagoras Theorem

Question 9 Exercise 13(A)

A parallelogram ABCD has P the mid-point of DC and Q a point of AC such that CQ =\frac{1}{4} \mathrm{AC} . \text { PQ produced meets } \mathrm{BC} \text { at } \mathrm{R} .

Prove that:

(i) R is the mid-point of BC,

\text { (ii) } \quad \mathrm{PR}=\frac{1}{2} \mathrm{DB}

Solution:

For help, we draw the diagonal BD as shown below

The diagonal AC and BD cuts at point X.

We know that the diagonal of a parallelogram intersect equally with each other. Therefore

AX=CX and BX=DX

Given,

\begin{array}{l} \mathrm{CQ}=\frac{1}{4} \mathrm{AC} \\ \mathrm{CQ}=\frac{1}{4} \times 2 \mathrm{CX} \\ \mathrm{CQ}=\frac{1}{2} \mathrm{CX} \end{array}

Therefore Q is the midpoint of CX.

(i) For triangle CDX PQ||DX or PR||BD

Since for triangle CBX

Q is the midpoint of CX and QR||BX. Therefore R is the midpoint of BC

(ii) For triangle BCD

\begin{aligned} &\text { As } \mathrm{P} \text { and } \mathrm{R} \text { are the midpoint of } \mathrm{CD} \text { and } \mathrm{BC} \text { , therefore }\\ &P R=\frac{1}{2} D B \end{aligned}

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