Question

In the given figure a circle touches the sides $PQ,QR$ and $PR$ of $△PQR$ at the points $X,Y$ and $Z$ respectively. Show that $PX+QY+RZ=XQ+YR+ZP=\frac{1}{2}$ (perimeter of $△PQR$).

Answer 1

Since lenth from an exterior point to a circle are equal.

Therefore PX$=$PZ.....1

QY$=$QX......2

RZ$=$RY........3

Adding equation 1,2 and 3 ,we get

PX+QY+RZ $=$ PZ+QX+RY

Now,

Perimeter of triangle PQR $=$PQ+PR+QR

$=$ (PX+XQ)+(QY+YR)+(PZ+ZR)

$=$ (PX+PZ)+(QX+QY)+(RZ+RY)

$=$ 2PX+2QY+2RZ

$=$ 2(PX+QY+RZ)

Therefore,PX+QY+RZ $=$ XQ+YR+ZP $=$$\frac{1}{2}$(perimeter of triangle PQR)

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