Question

By basic proportionality theorem In triangle XYZ if seg PQll seg XY then​

Answer 1

$\red{\underline{\underline{Answer:}}}$

$\sf{\frac{XP}{PY}=\frac{ZQ}{QY}}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{In \ \triangle \ XYZ \ and \ \triangle \ PYQ,}$

$\sf{seg \ PQ \ is \ parallel \ to \ seg \ XZ..... given}$

$\sf{\angle \ YXZ=\angle \ YPQ... corresponding \ angles}$

$\sf{\angle \ YZX=\angle \ YQP... corresponding \ angles}$

$\sf{\angle \ XYZ=\angle \ PYQ... common \ angle}$

$\sf{\therefore{By \ AAA \ test \ of \ similarity.}}$

$\sf{\triangle \ XYZ \ and \ \triangle \ PYQ \ are \ similar.}$

$\sf{\therefore{\frac{XY}{PY}=\frac{ZY}{QY}...(c.s.s.t.)}}$

$\sf{By \ dividendo}$

$\sf{\frac{XY-PY}{PY}=\frac{ZY-QY}{QY}}$

$\sf{But, \ X-P-Y \ and \ Z-Q-Y}$

$\sf{\therefore{XY-PY=XP \ and \ ZY-QY=ZQ}}$

$\sf\purple{\implies{\frac{XP}{PY}=\frac{ZQ}{QY}}}$

Answer 2

$\purple{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}$

$\rm{By \ basic \ proportionality \ theorem \ In \ triangle \ XYZ \ if \ seg \ PQ \ ll \ seg \ XY \ then}$

$\rm{a) \: \: \: \frac{AP}{BP} \: = \: \frac{QC}{BQ} }$

$\rm{b) \: \: \: \frac{XP}{PY} \: = \: \frac{YQ}{QZ} }$

$\rm{a) \: \: \: \frac{XP}{PY} \: = \: \frac{ZQ}{QY} }$

$\rm{a) \: \: \: \frac{XY}{YZ} \: = \: \frac{QZ}{YQ} }$

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$\purple{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}$

$\bf{\pink{\boxed{\star{\blue{c) \ \frac{XP}{PY}=\frac{ZQ}{QY}}}}}}$

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$\sf\large\underline\pink{Given:}$

$\rm{In \ triangle \ XYZ \ if \ seg \ PQ \ ll \ seg \ XY }$

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$\sf\large\underline\blue{To \ Find:}$

$\rm{Correct \ option}$

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$\green{\underline{\red{\boxed{\boxed{\purple{\star{\sf Solution:}}}}}}}$

$\rm{In \ \triangle \ XYZ \ and \ \triangle \ PYQ,}$

$\rm{Given \ that }$

$\rm{seg \ PQ \ is \ parallel \ to \ seg \ XZ}$

$\rm{\angle \ YXZ \ = \ \angle \ YPQ \ ...\ corresponding \ angles}$

$\rm{\angle \ YZX \ = \ \angle \ YQP \ .... \ corresponding \ angles}$

$\rm{\angle \ XYZ=\angle \ PYQ \ ..... \ common \ angle}$

$\rm{\implies{By \ AAA \ test \ of \ similarity.}}$

$\rm\therefore{\triangle \ XYZ \ and \ \triangle \ PYQ \ are \ similar.}$

$\rm{\implies{\frac{XY}{PY}=\frac{ZY}{QY}}}$

$\rm{By \ using \ dividendo \ property}$

$\rm{\frac{XY-PY}{PY}=\frac{ZY-QY}{QY}}$

$\rm{But, \ X-P-Y \ and \ Z-Q-Y}$

$\rm{\implies{XY-PY=XP} }$

$\rm{and \ ZY-QY=ZQ}$

$\sf{\therefore{\frac{XP}{PY} \ = \ \frac{ZQ}{QY}}}$

$\bf{\pink{\boxed{\star{\blue{c) \ \frac{XP}{PY}=\frac{ZQ}{QY}}}}}}$

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$\tt\red{Hope \ it \ helps \ :))}$