Question

The line segment PQ is parallel to side AC of triangle ABC and it divides the triangle into two equal parts. Find the ratio AP/AB.​

Answer 1

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Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{In \: \triangle \: ABC} \\ &\sf{PQ \: \parallel \: AC} \\ &\sf{ar( \triangle \: ABC) \: = 2 \: ar \: ( \triangle \: BPQ)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{\dfrac{AP}{AB} }  \end{cases}\end{gathered}\end{gathered}$

Concept Used

$\boxed{ \blue{\tt \:  Area \: Ratio \: Theorem }}$

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

$\large\underline\purple{\bold{Solution :- }}$

☆ Now, Consider ️ ABC and ️ PBQ

$\tt \:  ⟼ \angle \: BPQ \: = \angle \: BAC \: \: (corresponding \: angles)$

$\tt \:  ⟼ \angle \: ABC \: = \: \angle \: PBQ \: \: (commom)$

$\tt\implies \: \boxed{ \purple{ \tt \:  \triangle \: ABC \: \sim \: \triangle \: PBQ}}$

$\bf \:  \therefore \: by \: Area \: Ratio \: Theorem$

$\tt \:  \longrightarrow \: \dfrac{ar( \triangle \: PBQ)}{ar(\triangle \: ABC)} = \dfrac{ {PB}^{2} }{ {AB}^{2} }$

$\tt \:  \longrightarrow \: \dfrac{ar( \triangle \: PBQ)}{2 ar(\triangle \: PBQ)} = \dfrac{ {PB}^{2} }{ {AB}^{2} }$

$\tt \:  \longrightarrow \: \dfrac{1}{2} = \dfrac{ {PB}^{2} }{ {AB}^{2} }$

$\tt \:  ⟼ \dfrac{PB}{AB} = \dfrac{1}{ \sqrt{2} }$

$\tt \:  \longrightarrow \: \dfrac{AB - AP}{AB} = \dfrac{1}{ \sqrt{2} }$

$\tt \:  \longrightarrow \: \dfrac{AB}{AB} - \dfrac{AP}{AB} = \dfrac{1}{ \sqrt{2} }$

$\tt \:  \longrightarrow \: 1 - \dfrac{AP}{AB} = \dfrac{1}{ \sqrt{2} }$

$\tt \:  \longrightarrow \: \dfrac{AP}{AB} =1 - \dfrac{1}{ \sqrt{2} }$

$\tt\implies \boxed{ \bf \purple{\:\dfrac{AP}{AB} = \dfrac{ \sqrt{2} - 1 }{ \sqrt{2} }} }$