OP Malhotra Class 9 Maths Solutions Chapter 10 Pythagoras Theorem Exercise 10 (B)

S Chand Class 9 ICSE Maths Solutions Chapter 10 Pythagoras Theorem Ex 10(B)

Question 1. ABCD is a square, prove that \( AC^2 = 2AB^2 \).
Answer:
Given: ABCD is a square where AC is its diagonal.
To prove: \( AC^2 = 2AB^2 \).
Proof:
In triangle ABC, angle B is 90 degrees because it's a corner of a square. So, we can use the Pythagoras Theorem.
\( AC^2 = AB^2 + BC^2 \)
Since ABCD is a square, all sides are equal, so \( AB = BC \).
Substitute \( BC = AB \) into the equation:
\( AC^2 = AB^2 + AB^2 \)
\( AC^2 = 2AB^2 \)
Hence, it is proved that the square of the diagonal is twice the square of one side.
In simple words: For a square, if you draw a line from one corner to the opposite one (called the diagonal), its length squared is always double the length of one side squared. This is because a square has right angles, allowing Pythagoras's theorem to be used.

🎯 Exam Tip: Remember that in a square, all angles are 90 degrees and all sides are equal. These properties are key to applying the Pythagorean theorem correctly in such problems.

Question 2. In the figure, AB = BC = CA = 2a and segment AD ⊥ side BC. Draw that
(i) \( AD = a\sqrt{3} \)
(ii) area of \( \triangle ABC = a^2\sqrt{3} \)
Answer:
Given: In \( \triangle ABC \), \( AB = BC = CA = 2a \). Segment AD is perpendicular to side BC.
To prove:
(i) \( AD = a\sqrt{3} \)
(ii) area of \( \triangle ABC = a^2\sqrt{3} \)
Proof:
(i) The sides of \( \triangle ABC \) are all equal (each is \( 2a \)), which means it is an equilateral triangle.
In an equilateral triangle, the altitude (AD) also bisects the base (BC).
So, D is the midpoint of BC.
\( BD = DC = \frac{1}{2} BC = \frac{1}{2} (2a) = a \).
Now, consider the right-angled triangle ABD (since AD ⊥ BC).
By Pythagoras Theorem:
\( AB^2 = AD^2 + BD^2 \)
Substitute the known values:
\( (2a)^2 = AD^2 + (a)^2 \)
\( 4a^2 = AD^2 + a^2 \)
To find AD, rearrange the equation:
\( AD^2 = 4a^2 - a^2 \)
\( AD^2 = 3a^2 \)
Take the square root of both sides:
\( AD = \sqrt{3a^2} \)
\( AD = a\sqrt{3} \) (Hence proved for part i).
(ii) The area of a triangle is given by the formula: \( \frac{1}{2} \times \text{base} \times \text{altitude} \).
For \( \triangle ABC \), the base is BC and the altitude is AD.
Area of \( \triangle ABC = \frac{1}{2} \times BC \times AD \)
Substitute the values of BC and AD:
Area of \( \triangle ABC = \frac{1}{2} \times (2a) \times (a\sqrt{3}) \)
Area of \( \triangle ABC = a^2\sqrt{3} \) (Hence proved for part ii). This shows how the altitude helps calculate the area efficiently.
In simple words: This problem shows how to find the height and area of a special triangle where all sides are equal (an equilateral triangle). First, we split it into two right-angled triangles to find the height using the Pythagoras rule. Then, we use the height and base to calculate the total area.

🎯 Exam Tip: For equilateral triangles, always remember that the altitude from a vertex to the opposite side bisects that side. This forms two congruent right-angled triangles, making it easy to apply the Pythagorean theorem.

Question 3. In the figure, prove that \( AB^2 – AD^2 = CD^2 – CB^2 \).
Answer:
Given: In quadrilateral ABCD, \( \angle B = 90^\circ \) and \( \angle D = 90^\circ \).
To prove: \( AB^2 – AD^2 = CD^2 – CB^2 \).
Construction: Join AC, which acts as a common hypotenuse.
Proof:
Consider the right-angled triangle ABC (since \( \angle B = 90^\circ \)).
By Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \) ... (i)
Similarly, consider the right-angled triangle ADC (since \( \angle D = 90^\circ \)).
By Pythagoras Theorem:
\( AC^2 = AD^2 + CD^2 \) ... (ii)
From equations (i) and (ii), since both expressions are equal to \( AC^2 \), we can set them equal to each other:
\( AB^2 + BC^2 = AD^2 + CD^2 \)
Now, rearrange the terms to match the required proof:
\( AB^2 - AD^2 = CD^2 - BC^2 \)
Hence, it is proved that the difference of squares of sides is equal on both parts of the diagonal. This demonstrates how a common diagonal links two right triangles.
In simple words: We have a shape with two right angles. By drawing a line across it (a diagonal), we create two right-angled triangles. We use the Pythagoras rule for both triangles, and because they share the same diagonal, we can connect their equations to prove the given statement.

🎯 Exam Tip: When dealing with quadrilaterals that have two right angles, connecting the non-adjacent vertices creates a common hypotenuse. This allows you to use the Pythagorean theorem twice and then equate the expressions for the shared hypotenuse.

Question 4. In a \( \triangle ABC \), AD \( \perp \) BC. Prove that \( AB^2 + CD^2 = AC^2 + BD^2 \).
Answer:
Given: In \( \triangle ABC \), AD is perpendicular to BC (AD \( \perp \) BC).
To prove: \( AB^2 + CD^2 = AC^2 + BD^2 \).
Proof:
Consider the right-angled triangle ABD (since AD \( \perp \) BC).
By Pythagoras Theorem:
\( AB^2 = BD^2 + AD^2 \)
Rearrange this to find \( AD^2 \):
\( AD^2 = AB^2 - BD^2 \) ... (i)
Similarly, consider the right-angled triangle ACD (since AD \( \perp \) BC).
By Pythagoras Theorem:
\( AC^2 = AD^2 + CD^2 \)
Rearrange this to find \( AD^2 \):
\( AD^2 = AC^2 - CD^2 \) ... (ii)
From equations (i) and (ii), since both expressions are equal to \( AD^2 \), we can set them equal to each other:
\( AB^2 - BD^2 = AC^2 - CD^2 \)
Now, rearrange the terms to match the required proof:
\( AB^2 + CD^2 = AC^2 + BD^2 \)
Hence, it is proved. This illustrates how the common altitude relates the sides of two right triangles within a larger triangle.
In simple words: We have a big triangle with a line drawn straight down from the top to the bottom, making two smaller right-angled triangles. By using the Pythagoras rule for both small triangles to find the square of the height, we can show that a certain sum of squared sides on one side equals a similar sum on the other.

🎯 Exam Tip: When a perpendicular is drawn from a vertex to the opposite side, it creates two right-angled triangles. Focus on expressing the square of the common altitude in terms of the other sides for both triangles, and then equate these expressions.

Question 5. In a quadrilateral ABCD, the diagonals AC, BD intersect at right angles. Prove that \( AB^2 + CD^2 = BC^2 + DA^2 \).
[No marks will be given if it is assumed that ABCD is either a rhombus or a square].
Answer:
Given: In quadrilateral ABCD, diagonals AC and BD intersect each other at right angles at point O.
To prove: \( AB^2 + CD^2 = BC^2 + DA^2 \).
Proof:
Since the diagonals intersect at right angles, we have four right-angled triangles: \( \triangle AOB, \triangle BOC, \triangle COD, \triangle DOA \).
In right-angled \( \triangle AOB \):
By Pythagoras Theorem: \( AB^2 = AO^2 + BO^2 \) ... (i)
In right-angled \( \triangle BOC \):
By Pythagoras Theorem: \( BC^2 = BO^2 + CO^2 \) ... (ii)
In right-angled \( \triangle COD \):
By Pythagoras Theorem: \( CD^2 = CO^2 + DO^2 \) ... (iii)
In right-angled \( \triangle DOA \):
By Pythagoras Theorem: \( DA^2 = DO^2 + AO^2 \) ... (iv)
Now, let's add equation (i) and equation (iii):
\( AB^2 + CD^2 = (AO^2 + BO^2) + (CO^2 + DO^2) \) ... (v)
Next, let's add equation (ii) and equation (iv):
\( BC^2 + DA^2 = (BO^2 + CO^2) + (DO^2 + AO^2) \)
Rearranging the terms, we get:
\( BC^2 + DA^2 = AO^2 + BO^2 + CO^2 + DO^2 \) ... (vi)
From equations (v) and (vi), we can see that the right-hand sides are identical.
Therefore, \( AB^2 + CD^2 = BC^2 + DA^2 \).
Hence, it is proved. This shows a general property of quadrilaterals with perpendicular diagonals, not just special cases like rhombuses or squares.
In simple words: If the two lines drawn across a four-sided shape (diagonals) cross at a perfect right angle, then a special rule applies. If you square two opposite sides and add them, it will be the same as squaring the other two opposite sides and adding them. This works because the diagonals create four mini right-angled triangles inside the shape.

🎯 Exam Tip: The crucial information here is that the diagonals intersect at right angles. This immediately sets up four right-angled triangles. Apply the Pythagorean theorem to each, and then cleverly combine the resulting equations by adding them.

Question 6. In \( \triangle ABC \), \( \angle B = 90^\circ \) and D is mid-point of BC. Prove that
(i) \( AC^2 = AD^2 + 3CD^2 \)
(ii) \( BC^2 = 4 (AD^2 – AB^2) \)
Answer:
Given: In \( \triangle ABC \), \( \angle B = 90^\circ \). D is the midpoint of BC. AD is joined.
To prove:
(i) \( AC^2 = AD^2 + 3CD^2 \)
(ii) \( BC^2 = 4 (AD^2 – AB^2) \)
Proof:
Since D is the midpoint of BC, we have \( BD = DC = \frac{1}{2} BC \). This means \( BC = 2CD \).
(i) Consider the right-angled triangle ABC (since \( \angle B = 90^\circ \)).
By Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \)
Substitute \( BC = 2CD \):
\( AC^2 = AB^2 + (2CD)^2 \)
\( AC^2 = AB^2 + 4CD^2 \) ... (1)
Now, consider the right-angled triangle ABD (since \( \angle B = 90^\circ \)).
By Pythagoras Theorem:
\( AD^2 = AB^2 + BD^2 \)
Since \( BD = CD \), substitute \( BD = CD \):
\( AD^2 = AB^2 + CD^2 \)
From this, we can express \( AB^2 \):
\( AB^2 = AD^2 - CD^2 \) ... (2)
Substitute equation (2) into equation (1):
\( AC^2 = (AD^2 - CD^2) + 4CD^2 \)
\( AC^2 = AD^2 + 3CD^2 \) (Hence proved for part i). This shows how the median to the hypotenuse in a right triangle relates to other segments.
(ii) From equation (2), we have \( AD^2 = AB^2 + BD^2 \).
Rearrange this equation to isolate \( BD^2 \):
\( BD^2 = AD^2 - AB^2 \)
We know that \( BC = 2BD \), so \( BC^2 = (2BD)^2 = 4BD^2 \).
Substitute the expression for \( BD^2 \):
\( BC^2 = 4 (AD^2 - AB^2) \) (Hence proved for part ii).
In simple words: We have a triangle with a right angle, and a point D is exactly in the middle of one side. We use the Pythagoras rule twice – once for the big triangle and once for a smaller one. Then, we link these equations using the fact that D is a midpoint to prove the two given statements about the sides.

🎯 Exam Tip: When a midpoint is involved in a right-angled triangle, express the full side length in terms of the segment to the midpoint (e.g., \( BC = 2BD \)). This allows you to substitute and simplify equations effectively, often by applying the Pythagorean theorem multiple times.

Question 7. The side BC of a square ABCD is produced to any point E. Prove that \( AE^2 = 2BC \cdot BE + CE^2 \).
Answer:
Given: ABCD is a square. Side BC is extended to point E. EA is joined.
To prove: \( AE^2 = 2BC \cdot BE + CE^2 \).
Proof:
Since ABCD is a square, we know that AB is perpendicular to BC (\( AB \perp BC \)).
This means \( \angle ABE \) is a right angle (\( 90^\circ \)).
Consider the right-angled triangle ABE.
By Pythagoras Theorem:
\( AE^2 = AB^2 + BE^2 \)
Since ABCD is a square, all sides are equal: \( AB = BC = CD = DA \).
So, we can replace \( AB^2 \) with \( BC^2 \):
\( AE^2 = BC^2 + BE^2 \)
We also know that \( BE \) is made up of BC and CE, so \( BE = BC + CE \).
Substitute \( BE = BC + CE \) into the equation:
\( AE^2 = BC^2 + (BC + CE)^2 \)
Expand the term \( (BC + CE)^2 \) using the identity \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( AE^2 = BC^2 + (BC^2 + 2BC \cdot CE + CE^2) \)
Combine the \( BC^2 \) terms:
\( AE^2 = 2BC^2 + 2BC \cdot CE + CE^2 \)
We want to express this in terms of \( BE \). We know \( BE = BC + CE \). So, \( CE = BE - BC \).
Also, from \( BE = BC + CE \), we can see that \( 2BC \cdot BE = 2BC (BC + CE) = 2BC^2 + 2BC \cdot CE \).
Therefore, we can substitute \( 2BC \cdot BE \) back into our expanded equation:
\( AE^2 = (2BC^2 + 2BC \cdot CE) + CE^2 \)
\( AE^2 = 2BC \cdot BE + CE^2 \)
Hence, it is proved. This shows how extending a side of a square affects the diagonal from the opposite vertex to the extended point.
In simple words: Imagine a square. If you make one of its bottom sides longer to a new point E, and then draw a line from the top-left corner (A) to this new point E, we can find the square of this new line's length. It turns out to be equal to two times the original side length multiplied by the total extended length (BE), plus the square of the small extra piece (CE). This is proven by using the Pythagoras theorem on the right-angled triangle formed.

🎯 Exam Tip: When a side of a square (or rectangle) is extended, it often forms a right-angled triangle with the opposite vertex. Break down the extended segment (e.g., BE = BC + CE) and use the properties of the square (e.g., \( AB=BC \)) in conjunction with the Pythagorean theorem.

Question 8. ABCD is a rhombus. Prove that \( AC^2 + BD^2 = 4AB^2 \).
Answer:
Given: ABCD is a rhombus. Its diagonals AC and BD bisect each other at right angles at point O.
To prove: \( AC^2 + BD^2 = 4AB^2 \).
Proof:
In a rhombus, diagonals bisect each other at right angles. This means that \( \triangle AOB \) is a right-angled triangle with \( \angle AOB = 90^\circ \).
Also, O is the midpoint of AC and BD.
So, \( AO = \frac{1}{2} AC \) and \( BO = \frac{1}{2} BD \).
Consider the right-angled \( \triangle AOB \).
By Pythagoras Theorem:
\( AB^2 = AO^2 + BO^2 \)
Substitute \( AO = \frac{1}{2} AC \) and \( BO = \frac{1}{2} BD \):
\( AB^2 = \left(\frac{1}{2} AC\right)^2 + \left(\frac{1}{2} BD\right)^2 \)
\( AB^2 = \frac{1}{4} AC^2 + \frac{1}{4} BD^2 \)
Multiply the entire equation by 4 to clear the fractions:
\( 4AB^2 = AC^2 + BD^2 \)
Rearrange the terms to match the required proof:
\( AC^2 + BD^2 = 4AB^2 \)
Hence, it is proved. This special property holds true for all rhombuses.
In simple words: A rhombus is a diamond shape where all four sides are equal. Its diagonals (lines crossing through the middle) cut each other in half at a perfect right angle. If you square the length of each diagonal and add them, the result is four times the square of one side of the rhombus. We prove this using the Pythagoras rule on one of the four small right-angled triangles formed inside.

🎯 Exam Tip: The key properties of a rhombus are that all sides are equal and its diagonals bisect each other at right angles. Use these facts to set up a right-angled triangle involving a side and half of each diagonal, then apply the Pythagorean theorem.

Question 9. In the figure, \( \angle B \) of \( \triangle ABC \) is an acute angle and AD \( \perp \) BC. Prove that \( AC^2 = AB^2 + BC^2 – 2BC \cdot BD \).
Answer:
Given: In \( \triangle ABC \), \( \angle B \) is an acute angle. AD is perpendicular to BC (AD \( \perp \) BC).
To prove: \( AC^2 = AB^2 + BC^2 – 2BC \cdot BD \).
Proof:
Consider the right-angled triangle ABD (since AD \( \perp \) BC).
By Pythagoras Theorem:
\( AB^2 = AD^2 + BD^2 \)
Rearrange to find \( AD^2 \):
\( AD^2 = AB^2 - BD^2 \) ... (i)
Now, consider the right-angled triangle ADC (since AD \( \perp \) BC).
By Pythagoras Theorem:
\( AC^2 = AD^2 + DC^2 \)
From the figure, we can see that \( DC = BC - BD \).
Substitute this into the equation:
\( AC^2 = AD^2 + (BC - BD)^2 \)
Expand \( (BC - BD)^2 \) using the identity \( (a-b)^2 = a^2 - 2ab + b^2 \):
\( AC^2 = AD^2 + (BC^2 - 2BC \cdot BD + BD^2) \)
Now, substitute the expression for \( AD^2 \) from equation (i) into this equation:
\( AC^2 = (AB^2 - BD^2) + BC^2 - 2BC \cdot BD + BD^2 \)
Notice that \( -BD^2 \) and \( +BD^2 \) cancel each other out.
\( AC^2 = AB^2 + BC^2 - 2BC \cdot BD \)
Hence, it is proved. This is a common formula relating the sides of a triangle where one angle is acute and an altitude is drawn.
In simple words: When we have a triangle with an angle smaller than 90 degrees (acute angle), and we draw a straight line from the opposite corner down to the base, making a right angle, we can find a special relationship between the sides. It states that the square of the side opposite the acute angle is equal to the sum of the squares of the other two sides, minus two times the base multiplied by the segment of the base next to the acute angle. This uses the Pythagoras rule twice and algebraic substitution.

🎯 Exam Tip: This theorem (Apollonius's theorem or Projection theorem for acute angles) is a direct application of the Pythagorean theorem. Remember to define the segments clearly (e.g., \( DC = BC - BD \)) and use algebraic expansion for squared binomials.

Question 10. In a quadrilateral ABCD, \( \angle B = 90^\circ \) and \( AD^2 = AB^2 + BC^2 + CD^2 \). Prove that \( \angle ACD = 90^\circ \).
Answer:
Given: In quadrilateral ABCD, \( \angle B = 90^\circ \) and \( AD^2 = AB^2 + BC^2 + CD^2 \).
To prove: \( \angle ACD = 90^\circ \).
Proof:
First, consider the right-angled triangle ABC (since \( \angle B = 90^\circ \)).
By Pythagoras Theorem:
\( AC^2 = AB^2 + BC^2 \) ... (i)
Now, we are given the condition for the quadrilateral:
\( AD^2 = AB^2 + BC^2 + CD^2 \)
Substitute \( AB^2 + BC^2 \) with \( AC^2 \) from equation (i):
\( AD^2 = AC^2 + CD^2 \)
This equation \( AD^2 = AC^2 + CD^2 \) is the converse of the Pythagoras Theorem for \( \triangle ACD \).
The converse of Pythagoras Theorem states that if the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
In \( \triangle ACD \), the side opposite to \( AD \) is \( \angle ACD \).
Therefore, by the converse of Pythagoras Theorem, \( \angle ACD = 90^\circ \).
Hence, it is proved. This demonstrates how a given relationship between side lengths can imply a right angle.
In simple words: We have a four-sided shape where one corner is a right angle. We're also told a special rule about the squared lengths of its sides. By using the Pythagoras rule on the part of the shape with the given right angle, we can then show that another corner in the shape must also be a right angle. This uses the opposite idea of Pythagoras's theorem.

🎯 Exam Tip: When you need to prove an angle is \( 90^\circ \), think about using the converse of the Pythagorean theorem. Look for a way to manipulate the given equations to show that the square of one side is equal to the sum of the squares of the other two sides in the relevant triangle.

Question 11. ABC is a triangle right angled at A and p is the length of the perpendicular from A on BC. Show that
(i) pa = bc Hence deduce that
(ii) \( \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2} \)
Answer:
Given: In \( \triangle ABC \), \( \angle A = 90^\circ \). AD is perpendicular to BC (AD \( \perp \) BC).
Let \( AD = p, AB = c, AC = b \), and \( BC = a \).
To show:
(i) \( pa = bc \)
(ii) \( \frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2} \)
Proof:
(i) The area of \( \triangle ABC \) can be calculated in two ways:
Using base BC and altitude AD:
Area \( = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times a \times p \).
Using base AB and altitude AC (since \( \angle A = 90^\circ \), AB and AC are perpendicular to each other):
Area \( = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times c \times b \).
Since both expressions represent the area of the same triangle, they must be equal:
\( \frac{1}{2} pa = \frac{1}{2} bc \)
Multiply both sides by 2:
\( pa = bc \) (Hence proved for part i). This formula is a very useful property of right-angled triangles.
(ii) From part (i), we have \( pa = bc \).
Square both sides of this equation:
\( (pa)^2 = (bc)^2 \)
\( p^2 a^2 = b^2 c^2 \)
Now, from \( \triangle ABC \), since \( \angle A = 90^\circ \), by Pythagoras Theorem:
\( BC^2 = AB^2 + AC^2 \)
Substitute the side lengths with their labels:
\( a^2 = c^2 + b^2 \) ... (Equation from Pythagoras)
Substitute \( a^2 = b^2 + c^2 \) into the equation \( p^2 a^2 = b^2 c^2 \):
\( p^2 (b^2 + c^2) = b^2 c^2 \)
To get the required form, divide both sides by \( p^2 b^2 c^2 \):
\( \frac{p^2 (b^2 + c^2)}{p^2 b^2 c^2} = \frac{b^2 c^2}{p^2 b^2 c^2} \)
\( \frac{b^2 + c^2}{b^2 c^2} = \frac{1}{p^2} \)
Now, separate the fraction on the left side:
\( \frac{b^2}{b^2 c^2} + \frac{c^2}{b^2 c^2} = \frac{1}{p^2} \)
Simplify each term:
\( \frac{1}{c^2} + \frac{1}{b^2} = \frac{1}{p^2} \)
Rearrange the terms:
\( \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{c^2} \) (Hence proved for part ii). This reciprocal relationship is another elegant property.
In simple words: For a right-angled triangle, if you draw a line straight down from the right-angle corner to the longest side (hypotenuse), we can find two useful facts. First, the length of this height multiplied by the hypotenuse is the same as multiplying the other two sides. Second, if you take 1 divided by the square of the height, it's equal to 1 divided by the square of one side plus 1 divided by the square of the other side. Both are shown using area formulas and the Pythagoras rule.

🎯 Exam Tip: For right-angled triangles, the area formula can be used in multiple ways. Equating the area calculated using different bases and altitudes is a powerful technique to derive relationships between side lengths and altitudes. Also, don't forget the Pythagorean theorem for the main triangle.

Question 12. In the given figure, \( \angle B \) is acute and segment AD \( \perp \) side BC. Show that
(i) \( b^2 = h^2 + a^2 + x^2 – 2ax \)
(ii) \( b^2 = a^2 + c^2 – 2ax \)
Answer:
Given: In \( \triangle ABC \), \( \angle B \) is an acute angle. AD is perpendicular to BC (AD \( \perp \) BC).
Let \( AD = h, BD = x, AC = b, AB = c, BC = a \). From the figure, \( DC = BC - BD = a - x \).
To prove:
(i) \( b^2 = h^2 + a^2 + x^2 – 2ax \)
(ii) \( b^2 = a^2 + c^2 – 2ax \)
Proof:
(i) Consider the right-angled triangle ADC (since AD \( \perp \) BC).
By Pythagoras Theorem:
\( AC^2 = AD^2 + DC^2 \)
Substitute the given labels: \( b^2 = h^2 + (a - x)^2 \)
Expand \( (a - x)^2 \):
\( b^2 = h^2 + (a^2 - 2ax + x^2) \)
Rearrange the terms:
\( b^2 = h^2 + a^2 + x^2 – 2ax \) (Hence proved for part i). This expresses one side squared in terms of the altitude and base segments.
(ii) We need to show \( b^2 = a^2 + c^2 – 2ax \).
From part (i), we have \( b^2 = h^2 + a^2 + x^2 – 2ax \).
Now, consider the right-angled triangle ABD (since AD \( \perp \) BC).
By Pythagoras Theorem:
\( AB^2 = AD^2 + BD^2 \)
Substitute the given labels: \( c^2 = h^2 + x^2 \)
From this, we can express \( h^2 + x^2 = c^2 \).
Substitute \( h^2 + x^2 \) with \( c^2 \) into the equation from part (i):
\( b^2 = (h^2 + x^2) + a^2 – 2ax \)
\( b^2 = c^2 + a^2 – 2ax \)
Rearrange the terms:
\( b^2 = a^2 + c^2 – 2ax \) (Hence proved for part ii). This is a general form of the cosine rule for an acute angle.
In simple words: This problem explores the relationships between the sides of a triangle when one angle is sharp (acute) and a height line is drawn. By splitting the triangle into two right-angled parts and using the Pythagoras rule, we can show that the square of one side (b) can be written in two ways: first, using the height (h) and parts of the base (a and x); and second, using the other two sides (a and c) and parts of the base.

🎯 Exam Tip: This problem is a derivation of the Projection Theorem (or Law of Cosines for an acute angle). The key steps involve applying the Pythagorean theorem to both right-angled triangles formed by the altitude and then substituting expressions to eliminate common terms like \( h^2 \).