Question

prove that AP/PQ= AB+AC /BC​

Answer 1

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{PQ(AB+BC+CA)}{BC(PQ)}

BC(PQ)

PQ(AB+BC+CA)

Since, area of triangles with same base is equal to the proportion of their heights.

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

Area(PBC)

Area(ABC)

= \frac{AQ}{PQ}

PQ

AQ

\frac{AQ}{PQ}

PQ

AQ

= \frac{(AB+BC+CA)}{BC}

BC

(AB+BC+CA)

\frac{AP+PQ}{PQ}

PQ

AP+PQ

=\frac{AB+BC+CA}{BC}

BC

AB+BC+CA

Dividing the denominator to each numerator, we get

\frac{AP}{PQ}

PQ

AP

+1 =\frac{AB+CA}{BC}+1

BC

AB+CA

+1

Hence , we get the required result,

\frac{AP}{PQ}

PQ

AP

= \frac{AB+AC}{BC}

BC

AB+AC

other method-:-

see the second attachment

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