Skip to content

Practice: triangles, midsegments and loci

Practice: triangles, midsegments and loci

1

D\displaystyle D, E\displaystyle E and F\displaystyle F are the midpoints of the sides AB\displaystyle AB, BC\displaystyle BC and CA\displaystyle CA of the triangle ABC\displaystyle ABC.
Prove: △ADF≅△DBE\displaystyle \triangle ADF\cong\triangle DBE.

2

In the triangle ABC\displaystyle ABC, the medians AD\displaystyle AD, BE\displaystyle BE and CF\displaystyle CF meet at G\displaystyle G.
AD=15\displaystyle {AD}={15}, BE=18\displaystyle {BE}={18} and CG=8\displaystyle {CG}={8}.

a
Find AG\displaystyle AG.

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

    / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

    b
    Find GE\displaystyle GE.

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

      / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

      c
      Find the median CF\displaystyle CF.

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

        / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

        3

        In the triangle ABC\displaystyle ABC, the angle C\displaystyle C is right.
        M\displaystyle M is the midpoint of the hypotenuse AB\displaystyle AB, and D\displaystyle D the midpoint of AC\displaystyle AC.
        Prove: MD⊥AC\displaystyle MD\perp AC.

        a

        Explain in words why MA=MC\displaystyle {MA}={MC}, by the perpendicular bisector – a new proof that the median to the hypotenuse is half of it.

        4

        A(1,2)\displaystyle A(1,2) and B(7,4)\displaystyle B(7,4) are two points.

        a

        Write the equation of the locus of the points at equal distances from A\displaystyle A and from B\displaystyle B.

        A=(1,2)\displaystyle A=(1,2)
        B=(7,4)\displaystyle B=(7,4)

        b

        Find the point on the x\displaystyle x-axis at equal distances from A\displaystyle A and from B\displaystyle B.

          5

          The vertices of the triangle ABC\displaystyle ABC are A(−3,0)\displaystyle A(-3,0), B(3,0)\displaystyle B(3,0) and C(0,4)\displaystyle C(0,4).

          a

          Explain why the centres of the circumscribed and of the inscribed circle are both on the y\displaystyle y-axis.

          b
          The centre of the circumscribed circle is O(0,k)\displaystyle O(0,k), and OB=OC\displaystyle {OB}={OC}. Find k\displaystyle k.
            c
            Find the radius of the circumscribed circle.

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

              / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

              d
              The radius r\displaystyle r of the inscribed circle: the area of the triangle is 12r\displaystyle \frac{1}{2}r times its perimeter (cut the triangle into three triangles at the centre). Find r\displaystyle r.

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

                / – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

                6

                In the triangle ABC\displaystyle ABC, BE\displaystyle BE and CF\displaystyle CF are altitudes, and BE=CF\displaystyle {BE}={CF}.
                Prove: AB=AC\displaystyle {AB}={AC}.
                (A hint: in the triangles BCE\displaystyle BCE and CBF\displaystyle CBF, which side is opposite the right angle?)