Question
A $\triangle\text{ABC}$ is given. If lines are drawn through A, B, C, parallel respectively to the sides BC, CA and AB, forming $\triangle\text{PQR},$ as shown in the adjoining figure, show that $\text{BC}=\frac{1}{2}\text{QR}.$
✓
Answer
Given: A $\triangle\text{ABC}$ in which through points A, B and C, line QR, QP and RP are drawn parallel to BC, CA and AB.
To prove: $\text{BC}=\frac{1}{2}\text{QR}$
Proof: Since AR || BC and AB || RC [Given] So, ABCR is a parallelogram. Therefore AR = BC ...(i) Also, AQ || BC and QB || AC So, AQBC is a parallelogram.Therefore QA = BC ...(ii) Adding both side of (i) and (ii), we get AR + QA = BC + BC ⇒ QR = 2BC$\Rightarrow\text{BC}=\frac{\text{QR}}{2}$
$\therefore\text{BC}=\frac{1}{2}\text{QR}$
To prove: $\text{BC}=\frac{1}{2}\text{QR}$Proof: Since AR || BC and AB || RC [Given] So, ABCR is a parallelogram. Therefore AR = BC ...(i) Also, AQ || BC and QB || AC So, AQBC is a parallelogram.Therefore QA = BC ...(ii) Adding both side of (i) and (ii), we get AR + QA = BC + BC ⇒ QR = 2BC$\Rightarrow\text{BC}=\frac{\text{QR}}{2}$
$\therefore\text{BC}=\frac{1}{2}\text{QR}$
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