Question

pls answer guys I will follow the one answering properly !!!​

Answer 1

Consider, Area (ΔPBC) = $\frac{1}{2}$BC×PQ

Area (ΔABC) =$\frac{1}{2}$(AB+BC+CA)×PQ

$\frac{Area(ABC)}{Area(PBC)}$=$\frac{PQ(AB+ BC+CA)}{BC(PQ)}[\tex]</p><p></p><p><strong>Since, area of triangles with same base is equal to the proportion of their heights.</strong></p><p></p><p>[tex]\frac{Area(ABC)}{Area(PBC)}$=$\frac{AQ}{PQ}$

$\frac{AQ}{PQ}$=$\frac{(AB+ BC+CA)}{BC}[\tex]</p><p></p><p>[tex]\frac{AQ+PQ}{PQ}$=$\frac{(AB+ BC+CA)}{BC}[\tex]</p><p></p><p></p><p><strong><u>Dividing the denominator to each numerator, we get</u></strong></p><p></p><p></p><p>[tex]\frac{AQ}{PQ}$+1 = $\frac{AB+CA}{BC}[\tex] +1</p><p></p><h3><strong><u>Hence , we get the required result,</u></strong></h3><p></p><p>[tex]\frac{AQ}{PQ}$=[tex]\frac{AB+CA}{BC}[\tex]

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Answer 2

Answer:

Consider, Area (ΔPBC) = \frac{1}{2}

2

1

BC×PQ

Area (ΔABC) =\frac{1}{2}

2

1

(AB+BC+CA)×PQ

\frac{Area(ABC)}{Area(PBC)}

Area(PBC)

Area(ABC)

= =\frac{AQ}{PQ}

PQ

AQ

\frac{AQ}{PQ}

PQ

AQ

= =+1 = =[tex]\frac{AB+CA}{BC}[\tex]

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Step-by-step explanation:

hope its help you

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