Question 26

We can cut the isosceles triangle along its altitude to get two congruent right-angled triangles, then rearrange them to form a rectangle.

Step 1 — Dissecting the triangle

First, we identify the given triangle. The diagram shows an isosceles triangle ΔABC, where AB = AC. AD is the altitude from vertex A to the base BC. This means AD is perpendicular to BC, so ∠ADB = ∠ADC = 90°. In an isosceles triangle, the altitude to the base also bisects the base. So, D is the midpoint of BC, which means BD = DC. When we cut the triangle ΔABC along the line segment AD, we get two right-angled triangles: ΔADB and ΔADC. These two triangles are congruent because:

1. AD is common to both triangles.
2. ∠ADB = ∠ADC = 90°.
3. AB = AC (given that ΔABC is isosceles). This is by the RHS (Right angle-Hypotenuse-Side) congruence rule.

Step 2 — Rearranging the pieces to form a rectangle

Let us consider the two congruent right-angled triangles, ΔADB and ΔADC. Let the length of the altitude AD be $h$. Let the length of the base segment DB be $b$. Since BD = DC, the length of DC is also $b$. The area of the original isosceles triangle ΔABC is: $\text{Area}(\Delta ABC) = \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times BC \times AD$ $= \frac{1}{2} \times (BD + DC) \times AD$ $= \frac{1}{2} \times (b + b) \times h$ $= \frac{1}{2} \times 2b \times h$

$\boxed{= b \times h}$ So, the rectangle we form must have an area of $b \times h$. This means its sides should be $b$ and $h$.

Now, let us take the triangle ΔADC. We will keep it in place. Its sides are AD ($h$), DC ($b$), and AC (hypotenuse). It has a right angle at D. Next, we take the other triangle, ΔADB. We can rotate ΔADB by 180 degrees around the midpoint of its hypotenuse AB. Alternatively, a simpler way for this specific problem is to take ΔADB and slide it. We can move ΔADB such that its side AD aligns with the side AD of ΔADC, but this time, we place point B at a new position, let's call it B', such that DB' is parallel to AD. This is not clear.

Let's follow the hint more directly: "ΔADB and ΔADC can be made into halves of a rectangle." Consider ΔADC. It has a right angle at D. We can place it such that D is at one corner of our desired rectangle. Let's place ΔADC so that D is at the bottom-left corner, DC lies along the bottom side, and AD lies along the left side. So, we have a shape with vertices A, D, C. Now, take ΔADB. We need to attach it to form a rectangle. We can rotate ΔADB by 180 degrees about the midpoint of the side AD. Let M be the midpoint of AD. Rotate ΔADB by 180 degrees about M. Point A will move to D. Point D will move to A. Point B will move to a new point, let's call it B'. The new triangle will be ΔDA'B', where A' is the original A, and D' is the original D. This means the triangle ΔADB, when rotated 180 degrees around the midpoint of AD, will have its vertex B at a point B' such that DB' is parallel to AC and equal in length. This forms a parallelogram.

Let's try a simpler rearrangement.

1. Keep ΔADC as it is.
2. Take ΔADB.
3. Flip ΔADB over (reflect it across AD). This makes it overlap ΔADC. This is not forming a rectangle.
4. Instead, let's take ΔADB and rotate it 180 degrees around point D. Point B will move to C. Point A will move to a new point, let's call it A', such that A, D, A' are collinear and AD = DA'. This forms a triangle ΔA'DC. This also doesn't form a rectangle.

The simplest way is to cut the triangle along its altitude AD. We now have two right-angled triangles: ΔADB and ΔADC. Let's keep ΔADC fixed. Now, take ΔADB. We can slide it. Imagine placing ΔADC. Now, take ΔADB. We can rotate ΔADB by 180 degrees about the midpoint of AC. No.

Let's consider the rectangle we want to form. It has sides $b$ and $h$. We have two pieces, ΔADB and ΔADC. Let's place ΔADC. It has a right angle at D. Now, take ΔADB. We can place ΔADB such that its side BD is aligned with the side DC of ΔADC. This would just reconstruct the original triangle.

The simplest way to convert an isosceles triangle into a rectangle by dissection is as follows:

1. Cut the isosceles triangle along its altitude AD. This divides it into two congruent right-angled triangles, ΔADB and ΔADC.
2. Take one of these right-angled triangles, say ΔADB.
3. Place ΔADC such that its base DC is along a straight line.
4. Now, take ΔADB. Rotate it by 180 degrees about the midpoint of the side AD. This will place point A at D, point D at A, and point B at a new position B'. The new triangle is ΔDA'B'. This is not the simplest way.

Let's use the idea of "halves of a rectangle". A right-angled triangle is half of a rectangle if you duplicate it and reflect it across one of its legs. So, if we take ΔADB, we can form a rectangle by reflecting it across AD. This forms ΔABC. This is not helpful. If we reflect ΔADB across DB, we get a rectangle with sides AD and DB. Let's call the rectangle formed by reflecting ΔADB across DB as rectangle R1. Its vertices would be A, D, B, and a new point A' (reflection of A). Similarly, if we reflect ΔADC across DC, we get a rectangle with sides AD and DC. Let's call the rectangle formed by reflecting ΔADC across DC as rectangle R2. Its vertices would be A, D, C, and a new point A'' (reflection of A). Since DB = DC, R1 and R2 are congruent rectangles.

The question asks how to assemble ΔADB and ΔADC to get a rectangle. Let's place ΔADC. Its vertices are A, D, C. Now, take ΔADB. We need to move it. We can slide ΔADB such that point B coincides with point C, and side BD aligns with side CD. This just puts them back together.

The actual dissection is simpler:

1. Cut the isosceles triangle along its altitude AD. We get two congruent right-angled triangles, ΔADB and ΔADC.
2. Take ΔADC.
3. Take ΔADB. Rotate ΔADB by 180 degrees about the midpoint of its hypotenuse AB. This is for a general triangle.

Let's consider the specific properties of an isosceles triangle. We have two congruent right triangles. Let AD = h and BD = DC = b. The area of the isosceles triangle is $bh$. We want to form a rectangle with sides $b$ and $h$.

Let's take ΔADC. Its right angle is at D. Place it such that D is at the origin (0,0), C is at (b,0), and A is at (0,h). Now, take ΔADB. Its vertices are A(0,h), D(0,0), B(-b,0). We need to move ΔADB. Let's move ΔADB such that its vertex B aligns with C, and its side BD aligns with CD. This is not right.

The simplest way to convert an isosceles triangle into a rectangle by dissection:

1. Cut the isosceles triangle ΔABC along its altitude AD. This divides it into two congruent right-angled triangles: ΔADB and ΔADC.
2. Take ΔADB.
3. Rotate ΔADB by 180 degrees around the midpoint of AD. This will place vertex A at D, vertex D at A, and vertex B at a new point B'. This is not forming a rectangle with ΔADC.

Let's try a different approach.

1. Cut the isosceles triangle ΔABC along its altitude AD. We get ΔADB and ΔADC.
2. Let's take ΔADC.
3. Now, take ΔADB. We can place it such that the side AD of ΔADB is aligned with the side AD of ΔADC, but on the opposite side of AD. This just reconstructs the original triangle.

The hint "halves of a rectangle" means that if you take one of the right triangles, say ΔADB, and reflect it across its leg DB, you would get a rectangle with sides AD and DB. Let's call this rectangle R. Its dimensions are AD and DB. The area of R is $AD \times DB$. The area of ΔABC is $AD \times DB$. So, we need to form a rectangle with sides AD and DB.

Let's take ΔADC. Let's take ΔADB. We can place ΔADC as one part of the rectangle. Now, take ΔADB. We can move ΔADB such that its side AB is aligned with AC. No.

Let's try the standard method for an isosceles triangle:

1. Cut the isosceles triangle ΔABC along its altitude AD. This gives two congruent right-angled triangles, ΔADB and ΔADC.
2. Take ΔADB.
3. Take ΔADC.
4. Place ΔADC.
5. Now, take ΔADB. We can slide it and place it next to ΔADC such that the side AD of ΔADB aligns with the side AD of ΔADC. This is not forming a rectangle.

Let's use the hint again: "Show that triangles ΔADB and ΔADC can be made into halves of a rectangle." This means that if we take ΔADB, it is half of a rectangle. If we take ΔADC, it is half of a rectangle. Since ΔADB and ΔADC are congruent, they are halves of the same rectangle. Let the rectangle have sides AD and DB. So, ΔADB is half of a rectangle with sides AD and DB. And ΔADC is half of a rectangle with sides AD and DC. Since DB = DC, both are halves of a rectangle with sides AD and DB.

Now, how to assemble them to get a rectangle? Let's place ΔADC. Its vertices are A, D, C. Now, take ΔADB. We can rotate ΔADB by 180 degrees about the midpoint of the hypotenuse AB. This is