KindiakMath

Proving Exercises

Problem 1. The diagram below shows a quadrilateral ABCD with the midpoints P, Q, R, S of AB, BC, CD, DA respectively.

Show that PQRS forms a parallelogram.

(Click for Solution)

Solution. It suffices to show that PQ || RS and QR || SP. Since P is the midpoint of AB = BA and Q is the midpoint of BC, we have PQ || AC by the midpoint theorem. Since R is the midpoint of CD = DC and S is the midpoint of DA, we have RS || CA = AC by the midpoint theorem. Therefore, PQ || RS, as required.

Similarly, since Q is the midpoint of BC = CB and R is the midpoint of CD, we have QR || BD by the midpoint theorem. Since S is the midpoint of DA = AD and P is the midpoint of AB, we have SP || DB = BD by the midpoint theorem. Therefore, QR || SP, as required.

Alternate Solution. By the midpoint theorem,

\begin{aligned} \overrightarrow{OP} &= \textstyle \frac 12 (\overrightarrow{OA} + \overrightarrow{OB}), \\ \overrightarrow{OQ} &= \textstyle \frac 12 (\overrightarrow{OB} + \overrightarrow{OC}), \\ \overrightarrow{OR} &= \textstyle \frac 12 (\overrightarrow{OC} + \overrightarrow{OD}), \\ \overrightarrow{OS} &= \textstyle \frac 12 (\overrightarrow{OD} + \overrightarrow{OA}). \end{aligned}

Therefore,

\begin{aligned} \overrightarrow{PQ} &= \overrightarrow{OQ} - \overrightarrow{OP} \\ &= \textstyle \frac 12 (\overrightarrow{OB} + \overrightarrow{OC}) - \textstyle \frac 12 (\overrightarrow{OA} + \overrightarrow{OB}) \\ &= \textstyle \frac 12 (\overrightarrow{OC} - \overrightarrow{OA}) \\ &= \textstyle \frac 12 \overrightarrow{AC}. \end{aligned}

Similarly,

\overrightarrow{QR} = \frac 12 \overrightarrow{BD}, \quad \overrightarrow{SR} = \frac 12 \overrightarrow{AC}, \quad \overrightarrow{PS} = \frac 12 \overrightarrow{BD}.

Therefore,

\overrightarrow{PQ} = \overrightarrow{SR}\quad \text{and} \quad \overrightarrow{QR} = \overrightarrow{PS}

implies that PQ || RS and QR || SP, as required.

Problem 2. The diagram below shows a triangle ABC with midpoint D of BC.

Show that AB2 + AC2 = 2 · AD2 + 2 · BD2. This result is called Apollonius’ theorem.

(Click for Solution)

Solution. By the law of cosines,

\begin{aligned} AB^2 &= AD^2 + BD^2 - 2 \cdot AD \cdot BD \cdot \cos \angle ADB, \\ AC^2 &= AD^2 + CD^2 - 2 \cdot AD \cdot CD \cdot \cos \angle ADC. \end{aligned}

Since D is the midpoint of BC, BD = CD. Since adjacent angles on a straight line sum to 180^\circ,

\cos \angle ADC = \cos (180^\circ -\angle ADB) = -{\cos \angle ADB}.

Summing the equations,

\begin{aligned} AB^2 + AC^2 &= 2 \cdot AD^2 + BD^2 + CD^2 \\ &= 2 \cdot AD^2 + BD^2 + BD^2 \\ &= 2 \cdot AD^2 + 2 \cdot BD^2. \end{aligned}

Problem 3. The diagram below shows a triangle ABC with midpoints A’, B’, C’ of BC, CA, AB respectively.

Show that \displaystyle \frac{AP}{PA'} = \frac{BP}{PB'} = \frac{CP}{PC'} = 2.

(Click for Solution)

Solution. Let \mathcal A(\Delta) denote the area of the triangle \Delta. Since the midpoints divide the possible bases of the triangle into 2, we have

\begin{aligned} &\mathcal A(BPA') + \mathcal A(APC') + \mathcal A(BPC') \\ &\phantom{====..} = \mathcal A(APC') + \mathcal A(BPC') + \mathcal A(APB'). \end{aligned}

Therefore, \mathcal A(BPA') = \mathcal A(APB'). Furthermore,

\begin{aligned} \mathcal A(CPA') &= \mathcal A(BPA'), \\ \mathcal A(CPB') &= \mathcal A(APB'). \end{aligned}

Since areas are additive,

\begin{aligned} \mathcal A(APC) &= \mathcal A(APB') + \mathcal A(CPB') \\ &= 2\cdot\mathcal A(APB') \\ &= 2\cdot\mathcal A(BPA') \\ &= 2\cdot\mathcal A(CPA') \\ &= 2\cdot\mathcal A(A'PC). \end{aligned}

Since \Delta APC and \Delta CPA' have the same height, we must have AP/PA' = 2/1 = 2. The other ratios follow similarly.

—Joel Kindiak, 31 Mar 26, 2359H

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