A quadrilateral is inscribed in a circle. If is a diameter, what can you say about and ? Explain your reasoning.

Understand the Question Two triangles with the same base length and the same perpendicular height have equal areas: . Points and trisect the base of , so the segments are equal in length: . Both the blue triangle and the red triangle share vertex , meaning they have the exact same perpendicular height drawn from to line . Because their bases and heights are identical, their areas are equal. Part (i) Show that the areas of the shaded blue triangle and the shaded red triangle are equal. Step 1 · Compare Areas of and Let points and on side trisect , such that: Let be the perpendicular height from vertex to the side . Area of blue triangle Area of red triangle Since both areas evaluate to , they are equal: Answer > (i) Part (ii) Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle. Step 1 · Dissection and Rearrangement Since , line segment is a median of : Similarly, since , line segment is a median of : Thus: To dissect to cover : 1. Draw a line through parallel to intersecting at point . 2. Cut along line segment to produce pieces and . 3. Rearrange the resulting pieces by translating them across the intermediate area of to completely cover . Answer > (ii) Cut along (where intersects at ) and rearrange the pieces to cover . Common Mistakes Assuming Heights Differ by Shape: Thinking that because and lean in different directions, their heights must differ. Since both share vertex and lie along the same line , their perpendicular heights are identical. Dissection by Simple Rotation: Assuming equiareal triangles can always be superimposed via rigid rotation without cutting; a dissection along parallel reference lines is necessary when side lengths and angles differ.

Circles and Geometric Shapes | EOT

Question 20

A quadrilateral $MNOP$ is inscribed in a circle. If $MN$ is a diameter, what can you say about $\angle MOP$ and $\angle MNP$ ? Explain your reasoning.

Solution

Understand the Question

Two triangles with the same base length and the same perpendicular height have equal areas: $\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}$ .
Points $D$ and $E$ trisect the base $BC$ of $\Delta ABC$ , so the segments are equal in length: $BD = DE = EC = x$ .
Both the blue triangle $\Delta ABD$ and the red triangle $\Delta AEC$ share vertex $A$ , meaning they have the exact same perpendicular height $h$ drawn from $A$ to line $BC$ .
Because their bases and heights are identical, their areas are equal.

(i) Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

Step 1 · Compare Areas of $\Delta ABD$ and $\Delta AEC$

Let points $D$ and $E$ on side $BC$ trisect $BC$ , such that: $BD = DE = EC = x$

Let $h$ be the perpendicular height from vertex $A$ to the side $BC$ . Diagram 1

Area of blue triangle $ABD$

\begin{aligned} \text{Area}(ABD) &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times BD \times h \\[0.6em] &= \dfrac{1}{2} \times x \times h \end{aligned}

Area of red triangle $AEC$

\begin{aligned} \text{Area}(AEC) &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times EC \times h \\[0.6em] &= \dfrac{1}{2} \times x \times h \end{aligned}

Since both areas evaluate to $\dfrac{1}{2} x h$ , they are equal: $\text{Area}(ABD) = \text{Area}(AEC)$

Answer

(i) $\text{Area}(ABD) = \text{Area}(AEC)$

(ii) Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Step 1 · Dissection and Rearrangement

Since $BD = DE$ , line segment $AD$ is a median of $\Delta ABE$ : $\text{Area}(ABD) = \text{Area}(ADE)$

Similarly, since $DE = EC$ , line segment $AE$ is a median of $\Delta ADC$ : $\text{Area}(ADE) = \text{Area}(AEC)$

Thus: $\text{Area}(ABD) = \text{Area}(ADE) = \text{Area}(AEC)$

To dissect $\Delta ABD$ to cover $\Delta AEC$ :

Draw a line through $D$ parallel to $AC$ intersecting $AE$ at point $P$ .
Cut $\Delta ABD$ along line segment $DP$ to produce pieces $\Delta ADP$ and $\Delta BDP$ .
Rearrange the resulting pieces by translating them across the intermediate area of $\Delta ADE$ to completely cover $\Delta AEC$ .

Answer

(ii) Cut $\Delta ABD$ along $DP$ (where $DP \parallel AC$ intersects $AE$ at $P$ ) and rearrange the pieces to cover $\Delta AEC$ .

Common Mistakes

Assuming Heights Differ by Shape: Thinking that because $\Delta ABD$ and $\Delta AEC$ lean in different directions, their heights must differ. Since both share vertex $A$ and lie along the same line $BC$ , their perpendicular heights are identical.
Dissection by Simple Rotation: Assuming equiareal triangles can always be superimposed via rigid rotation without cutting; a dissection along parallel reference lines is necessary when side lengths and angles differ.

Question 20

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