The midpoint theorem questions with answers

Questions

1. Question 1The midpoints of two sides of a triangle

   In triangle $PQR$, $X$ is the mid-point of $PQ$ and $Y$ is the mid-point of $PR$. $\angle PQR = 53^\circ$. Show that $XY$ is parallel to $QR$ and $XY = \dfrac{1}{2}QR$. Hence find $\angle PXY$.

   Show the worked solution

   1. Step 1: X and Y are the midpoints of two sides of triangle PQR$PX = XQ$ and $PY = YR$ (given)$XY \parallel QR$ and $XY = \dfrac{1}{2}QR$ (midpoint theorem)
   2. Step 2: XY ∥ QR, and PQ crosses both$\angle PXY = \angle PQR$ (corr. ∠s (XY ∥ QR))$\angle PXY = 53^\circ$
2. Question 2The midpoints of two sides of a triangle

   In triangle $PQR$, $X$ is the mid-point of $PQ$ and $Y$ is the mid-point of $PR$. $\angle PQR = 48^\circ$. Show that $XY$ is parallel to $QR$ and $XY = \dfrac{1}{2}QR$. Hence find $\angle PXY$.

   Show the worked solution

   1. Step 1: X and Y are the midpoints of two sides of triangle PQR$PX = XQ$ and $PY = YR$ (given)$XY \parallel QR$ and $XY = \dfrac{1}{2}QR$ (midpoint theorem)
   2. Step 2: XY ∥ QR, and PQ crosses both$\angle PXY = \angle PQR$ (corr. ∠s (XY ∥ QR))$\angle PXY = 48^\circ$
3. Question 3Midpoints of a quadrilateral: prove a parallelogram

   $ABCD$ is a quadrilateral. $P$, $Q$, $R$ and $S$ are the mid-points of $AB$, $BC$, $CD$ and $DA$. Prove that $PQRS$ is a parallelogram. Given that $AC = 11$ cm, find $PQ$.

   Show the worked solution

   1. Step 1: Join AC. In triangle ABC, P and Q are midpoints$PQ \parallel AC$ and $PQ = \dfrac{1}{2}AC$ (midpoint theorem)
   2. Step 2: In triangle ADC, S and R are midpoints$SR \parallel AC$ and $SR = \dfrac{1}{2}AC$ (midpoint theorem)
   3. Step 3: One pair of opposite sides is equal and parallelSo $PQ \parallel SR$ and $PQ = SR$.$PQRS$ is a parallelogram.
   4. Step 4: Use the result for PQ$\mathstrut PQ = \dfrac{1}{2} \times 11$$PQ = 5.5$ cm
4. Question 4Midpoints of a quadrilateral: prove a parallelogram

   $WXYZ$ is a quadrilateral. $J$, $K$, $L$ and $M$ are the mid-points of $WX$, $XY$, $YZ$ and $ZW$. Prove that $JKLM$ is a parallelogram. Given that $XZ = 10$ cm, find $KL$.

   Show the worked solution

   1. Step 1: Join WY. In triangle WXY, J and K are midpoints$JK \parallel WY$ and $JK = \dfrac{1}{2}WY$ (midpoint theorem)
   2. Step 2: In triangle WZY, M and L are midpoints$ML \parallel WY$ and $ML = \dfrac{1}{2}WY$ (midpoint theorem)
   3. Step 3: One pair of opposite sides is equal and parallelSo $JK \parallel ML$ and $JK = ML$.$JKLM$ is a parallelogram.
   4. Step 4: Join XZ. In triangle XYZ, K and L are midpoints$KL \parallel XZ$ and $KL = \dfrac{1}{2}XZ$ (midpoint theorem)In the same way, in triangle $XWZ$, $MJ = \dfrac{1}{2}XZ$.
   5. Step 5: Use the result for KL$\mathstrut KL = \dfrac{1}{2} \times 10$$KL = 5$ cm
5. Question 5Which special quadrilateral?

   In the quadrilateral $WXYZ$, the diagonals $WY$ and $XZ$ are equal in length. $J$, $K$, $L$ and $M$ are the mid-points of $WX$, $XY$, $YZ$ and $ZW$. Prove that $JKLM$ is a parallelogram, and state the most specific name for the quadrilateral $JKLM$.

   Show the worked solution

   1. Step 1: Join WY. In triangle WXY, J and K are midpoints$JK \parallel WY$ and $JK = \dfrac{1}{2}WY$ (midpoint theorem)
   2. Step 2: In triangle WZY, M and L are midpoints$ML \parallel WY$ and $ML = \dfrac{1}{2}WY$ (midpoint theorem)
   3. Step 3: One pair of opposite sides is equal and parallelSo $JK \parallel ML$ and $JK = ML$.$JKLM$ is a parallelogram.
   4. Step 4: Join XZ. In triangle XYZ, K and L are midpoints$KL \parallel XZ$ and $KL = \dfrac{1}{2}XZ$ (midpoint theorem)
   5. Step 5: What do you know about the diagonals of WXYZ?The diagonals are equal (given).
   6. Step 6: JK ∥ WY and KL ∥ XZ, each half as longSo the adjacent sides $JK$ and $KL$ are equal: $JKLM$ is a rhombus.
6. Question 6Which special quadrilateral?

   In the quadrilateral $EFGH$, the diagonals $EG$ and $FH$ are equal in length. $K$, $L$, $M$ and $N$ are the mid-points of $EF$, $FG$, $GH$ and $HE$. Prove that $KLMN$ is a parallelogram, and state the most specific name for the quadrilateral $KLMN$.

   Show the worked solution

   1. Step 1: Join EG. In triangle EFG, K and L are midpoints$KL \parallel EG$ and $KL = \dfrac{1}{2}EG$ (midpoint theorem)
   2. Step 2: In triangle EHG, N and M are midpoints$NM \parallel EG$ and $NM = \dfrac{1}{2}EG$ (midpoint theorem)
   3. Step 3: One pair of opposite sides is equal and parallelSo $KL \parallel NM$ and $KL = NM$.$KLMN$ is a parallelogram.
   4. Step 4: Join FH. In triangle FGH, L and M are midpoints$LM \parallel FH$ and $LM = \dfrac{1}{2}FH$ (midpoint theorem)
   5. Step 5: What do you know about the diagonals of EFGH?The diagonals are equal (given).
   6. Step 6: KL ∥ EG and LM ∥ FH, each half as longSo the adjacent sides $KL$ and $LM$ are equal: $KLMN$ is a rhombus.