ABC is a triangle and PQ is a Straight line meeting AB in P and AC in Q. If $A P = 1$ cm and $B P = 3$ cm, $A Q = 1.5$ cm, $C Q = 4.5$ cm.Prove $△ A B C \sim △ A P Q$

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Solution

Given In $△ A B C, A P = 1 c m, B P = 3 c m, A Q = 1.5 c m, C Q = 4.5 c m$

$\frac{A P}{B P} = \frac{1}{3}$

$\frac{A Q}{C Q} = \frac{1.5}{4.5} = \frac{1}{3}$

$∴ \frac{A P}{B P} = \frac{A Q}{C Q}$

Hence $P Q ∥ B C$ .
$∠ A$ is common in $△ A B C$ and $△ A P Q$ .

Hence $△ A P Q \sim △ A B C$ .

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ABC is a triangle and PQ is a Straight line meeting AB in P and AC in Q. If AP=1cm and BP=3cm, AQ=1.5cm, CQ=4.5cm.Prove △ABC∼△APQ

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