prove that AP/PQ= AB+AC /BC
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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
pls mark as brainest as it is so difficult to type such long
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