geometry problem please help me?

a circle touches the sides PQ , QR and PR of triangle PQR at the points X, Y and Z respectively. Show that PX + QY + RZ = XQ + YR + ZP = 1/2 (perimeter of triangle PQR)

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  • 8 years ago
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    i) As the circle touches PQ at X and PR at Z; PX = PZ ------------ (1)

    [Tangents from the same exterior point to the same circle are equal in measure]

    Similarly QY = QX ----------- (2) and

    RZ = RY ------------ (3)

    ii) Adding (1), (2) & (3),

    PX + QY + RZ = PZ + QX + RY ---------------(4) [Proved]

    iii) By geometrical addition, PX + XQ = PQ

    QY + YR = QR and RZ + ZP = RP

    Adding all these 3, PX + XQ + QY + YR + RZ + ZP = PQ + QR + RP

    Grouping them conveniently,

    (PX + QY + RZ) + (XQ + YR + ZP) = PQ + QR + RP = Perimeter of the triangle ------ (5)

    Hence from (4) & (5),

    2(PX + QY + RZ) = Perimeter of the triangle

    ==> PX + QY + RZ = (1/2)(Perimeter of the triangle PQR) ----------- (6)

    Thus from (4) & (6),

    PX +QY + RZ = XQ + YR + ZP = (1/2)(Perimeter of triangle PQR) [Proved]

  • 8 years ago

    Draw the radius from the circle to X, Y, Z. We know this is perpendicular because the radius of a circle is perpendicular to its tangent. Connect a line from the center of the circle to the vertices. By hypotenuse leg congruency PXC ~ PZC, RZC ~ RYC, QYC ~ QXC. This means PX=XQ, QY=YR, RZ=ZP. This means PX+QY+RZ=XQ+YR+ZP and since those line segments make up the perimeter and are equal PX+QY+RZ=XQ+YR+ZP=1/2 Perimeter (PQR)

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