In , PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides into two parts equal in area. Find BP AB .

We have, PQ || BC And ar ( ) = ar(trap. PQCB) ( )=ar( )-ar( ) 2ar( )=ar( ) ...(i) In and = [Common] = [Corresponding angles] Then, [By AA similarity] By area of similar triangle theorem ar( ) ar( ) = AP^2 AB^2 ar( ) ar( ) = AP^2 AB^2 [By using (i)] 1 2 = AP^2 AB^2 1 2 = AP AB^2 1 2 = AP AB [Taking square root] 1 2 = AB-BP AB 1 2 = AB AB - BP AB 1 2 =1- BP AB BP AB =1-1 2 BP AB = 2-1 2

In , PQ is a line segment intersecting AB at P and AC at Q such t…

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Question

In $\triangle\text{ABC, PQ}$ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides $\triangle\text{ABC}$ into two parts equal in area. Find $\frac{\text{BP}}{\text{AB}}.$

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Answer

We have,
PQ || BC
And ar $(\triangle\text{APQ})$ = ar(trap. PQCB)
$\Rightarrow\text{ar}(\triangle\text{APQ})=\text{ar}(\triangle\text{ABC})-\text{ar}(\triangle\text{APQ})$
$\Rightarrow\text{2ar}(\triangle\text{APQ})=\text{ar}(\triangle\text{ABC})\ \ ...(\text{i})$
In $\triangle\text{APQ and }\triangle\text{ABC}$
$\angle\text{A}=\angle\text{A}$ [Common]
$\angle\text{APQ}=\angle\text{B}$ [Corresponding angles]
Then, $\triangle\text{APQ}\sim\triangle\text{ABC}$ [By AA similarity]
$\therefore$ By area of similar triangle theorem
$\frac{\text{ar}(\triangle\text{APQ})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{AP}^2}{\text{AB}^2}$
$\Rightarrow\frac{\text{ar}(\triangle\text{APQ})}{\text{ar}(\triangle\text{APQ})}=\frac{\text{AP}^2}{\text{AB}^2}\ \ [\text{By using (i)}]$
$\Rightarrow\frac{1}{2}=\frac{\text{AP}^2}{\text{AB}^2}$
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AP}}{\text{AB}^2}$
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AP}}{\text{AB}}$ [Taking square root]
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AB}-\text{BP}}{\text{AB}}$
$\Rightarrow\frac{1}{\sqrt{2}}=\frac{\text{AB}}{\text{AB}}-\frac{\text{BP}}{\text{AB}}$
$\Rightarrow\frac{1}{\sqrt{2}}=1-\frac{\text{BP}}{\text{AB}}$
$\Rightarrow\frac{\text{BP}}{\text{AB}}=1-\frac{1}{\sqrt{2}}$
$\Rightarrow\frac{\text{BP}}{\text{AB}}=\frac{\sqrt{2}-1}{\sqrt{2}}$