Question

bisectors of angle b and angle c in triangle ABC meet each other at p. line AP cuts the side bc at q prove thatcAP/PQ =AB+AC/ BC

Answer 1

Answer:

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

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

$\frac{Area(ABC)}{Area(PBC)}$= $\frac{PQ(AB+BC+CA)}{BC(PQ)}$

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

$\frac{Area(ABC)}{Area(PBC)}$ = $\frac{AQ}{PQ}$

$\frac{AQ}{PQ}$ = $\frac{(AB+BC+CA)}{BC}$

$\frac{AP+PQ}{PQ}$=$\frac{AB+BC+CA}{BC}$

Dividing the denominator to each numerator, we get

$\frac{AP}{PQ}$+1 =$\frac{AB+CA}{BC}+1$

Hence , we get the required result,

$\frac{AP}{PQ}$ = $\frac{AB+AC}{BC}$

Answer 2

Answer:

Step-by-step explanation:

Prove that-AP/PQ=Ab+Ac/BC

In triangle ABQ,

Ray Bp Bisects angle ABQ

By the property of angle bisector

AB /BQ = AP /PQ... 1

In triangle ACQ,

Ray Cp bisects angle ACQ

Ac/ CQ =AP / PQ... 2

Ab/BQ = AC /CQ = AP /PQ...(from 1,2)

By theorem on equal ratios

AB + AC /BQ + CQ = AP / PQ

AB + AC / BC = AP / PQ

AP /PQ = AP /PQ... (BQ+QC =BC)

.....hence proved