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Question

A $$\triangle ABC$$ is given. If lines are drawn through $$AB,B,C$$ parallel respectively to the sides $$BC,CA$$ and $$AB$$ forming $$\triangle PQR$$, as shown in the figure, show that $$BC=\cfrac{1}{2}QR$$
1715714_89bb3a7bc5b14350af286e69d7f1caa4.PNG


Solution

We know that $$AR\parallel BC$$ and $$AB\parallel RC$$

From the figure we know that $$ABCR$$ is a parallelogram

So we get

$$AR=BC$$.... (1)

We know that $$AQ\parallel BC$$ and $$QB\parallel AC$$

From the figure we know  that $$AQBC$$ is a parallelogram

so we get

$$AQ=BC$$.......(2)

By adding both the equations

$$AR+QA=BC+BC$$

We know that $$AR+QA=QR$$

so we get

$$QR=2BC$$

dividing 2

$$BC=\dfrac{QR}{2}$$

$$BC=\dfrac{1}{2} QR$$

therefore it is proved that $$BC=\dfrac{1}{2}QR$$


Mathematics
RS Agarwal
Standard IX

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