EXAMPLES
Triangles • 7 Questions
Question 1
Hint available
ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig. 6.14). Show that AE BF ED FC .
Key Idea
When a set of parallel lines intersect two transversals, the intercepted segments on the transversals are proportional (Basic Proportionality Theorem / Thales theorem). Here AB, EF and DC are three parallel lines intersected by the transversals AD and BC.
Step-by-Step Solution
1. Identify the parallel lines and transversals\
The three lines AB, EF and DC are parallel. The non‑parallel sides AD and BC act as transversals intersecting these parallel lines at the points:
- AD meets AB at A, EF at E and DC at D.
- BC meets AB at B, EF at F and DC at C.
2. Consider the two triangles formed by the transversals\
- Triangle \(\triangle AED\) is formed by the points where AD meets the three parallel lines.
- Triangle \(\triangle BFC\) is formed by the points where BC meets the same three parallel lines.
3. Show that the two triangles are similar\
- \(\angle AED\) and \(\angle BFC\) are both straight angles (180°) on the same line, hence they are equal.
- Since AB \(||\) DC, the angle made by AD with AB equals the angle made by AD with DC. Therefore \(\angle ADE = \angle BCF\) (alternate interior angles).
- Similarly, because AB \(||\) EF, the angle made by AD with AB equals the angle made by AD with EF, giving \(\angle AED = \angle BFC\) again.
- With two corresponding angles equal, \(\triangle AED \sim \triangle BFC\) (AA similarity).
4. Write the proportion from the similarity\
From \(\triangle AED \sim \triangle BFC\), the ratios of corresponding sides are equal:
$$\frac{AE}{ED} = \frac{BF}{FC}.$$
5. Conclusion\
Hence, the line segment \(EF\) drawn parallel to the bases of the trapezium divides the non‑parallel sides proportionally, i.e. \(AE : ED = BF : FC\).
The three lines AB, EF and DC are parallel. The non‑parallel sides AD and BC act as transversals intersecting these parallel lines at the points:
- AD meets AB at A, EF at E and DC at D.
- BC meets AB at B, EF at F and DC at C.
2. Consider the two triangles formed by the transversals\
- Triangle \(\triangle AED\) is formed by the points where AD meets the three parallel lines.
- Triangle \(\triangle BFC\) is formed by the points where BC meets the same three parallel lines.
3. Show that the two triangles are similar\
- \(\angle AED\) and \(\angle BFC\) are both straight angles (180°) on the same line, hence they are equal.
- Since AB \(||\) DC, the angle made by AD with AB equals the angle made by AD with DC. Therefore \(\angle ADE = \angle BCF\) (alternate interior angles).
- Similarly, because AB \(||\) EF, the angle made by AD with AB equals the angle made by AD with EF, giving \(\angle AED = \angle BFC\) again.
- With two corresponding angles equal, \(\triangle AED \sim \triangle BFC\) (AA similarity).
4. Write the proportion from the similarity\
From \(\triangle AED \sim \triangle BFC\), the ratios of corresponding sides are equal:
$$\frac{AE}{ED} = \frac{BF}{FC}.$$
5. Conclusion\
Hence, the line segment \(EF\) drawn parallel to the bases of the trapezium divides the non‑parallel sides proportionally, i.e. \(AE : ED = BF : FC\).
Question 2
Hint available
In Fig. 6.16, PS SQ = PT TR and PST = PRQ. Prove that PQR is an isosceles triangle.
Key Idea
Use the SAS similarity criterion for triangles. If two triangles have an equal included angle and the products of the two sides containing the angle are equal, the triangles are similar. From the similarity of ΔPST and ΔPRQ we obtain the proportionality of corresponding sides, which leads to PQ = PR, showing that ΔPQR is isosceles.
Step-by-Step Solution
1. Given data
- $PS\cdot SQ = PT\cdot TR$.
- $\angle PST = \angle PRQ$.
- Points $S$ and $T$ lie on $PQ$ and $PR$ respectively (as shown in the figure).
2. Apply SAS similarity
- In triangles $\triangle PST$ and $\triangle PRQ$, the angle $\angle PST$ is equal to $\angle PRQ$.
- The products of the two sides enclosing these angles are equal: $PS\cdot SQ = PT\cdot TR$.
- Hence, by the SAS similarity criterion, \[\triangle PST \sim \triangle PRQ.\]
3. Correspondence of sides
- From the similarity, the following proportion holds:
$$\frac{PS}{PR}=\frac{PT}{PQ}=\frac{ST}{RQ}.$$
- Taking the first two ratios, we have
$$\frac{PS}{PR}=\frac{PT}{PQ}\quad\Rightarrow\quad PS\cdot PQ = PT\cdot PR.\]
4. Use the given product relation
- The given relation $PS\cdot SQ = PT\cdot TR$ can be rewritten as
$$\frac{PS}{PT}=\frac{TR}{SQ}.$$
- Since $S$ lies on $PQ$ and $T$ lies on $PR$, we have $SQ = PQ-PS$ and $TR = PR-PT$.
- Substituting these in the above proportion and simplifying yields
$$PS\cdot PQ = PT\cdot PR,$$
which is exactly the relation obtained from similarity.
5. Conclude equality of the two sides of $\triangle PQR$
- From the proportion $\frac{PT}{PQ}=\frac{PS}{PR}$ and the equality $PS\cdot PQ = PT\cdot PR$, we deduce
$$PQ = PR.$$
- Therefore, the sides $PQ$ and $PR$ of $\triangle PQR$ are equal.
6. Result
- Since two sides of $\triangle PQR$ are equal, $\triangle PQR$ is an isosceles triangle (with $PQ = PR$).
- $PS\cdot SQ = PT\cdot TR$.
- $\angle PST = \angle PRQ$.
- Points $S$ and $T$ lie on $PQ$ and $PR$ respectively (as shown in the figure).
2. Apply SAS similarity
- In triangles $\triangle PST$ and $\triangle PRQ$, the angle $\angle PST$ is equal to $\angle PRQ$.
- The products of the two sides enclosing these angles are equal: $PS\cdot SQ = PT\cdot TR$.
- Hence, by the SAS similarity criterion, \[\triangle PST \sim \triangle PRQ.\]
3. Correspondence of sides
- From the similarity, the following proportion holds:
$$\frac{PS}{PR}=\frac{PT}{PQ}=\frac{ST}{RQ}.$$
- Taking the first two ratios, we have
$$\frac{PS}{PR}=\frac{PT}{PQ}\quad\Rightarrow\quad PS\cdot PQ = PT\cdot PR.\]
4. Use the given product relation
- The given relation $PS\cdot SQ = PT\cdot TR$ can be rewritten as
$$\frac{PS}{PT}=\frac{TR}{SQ}.$$
- Since $S$ lies on $PQ$ and $T$ lies on $PR$, we have $SQ = PQ-PS$ and $TR = PR-PT$.
- Substituting these in the above proportion and simplifying yields
$$PS\cdot PQ = PT\cdot PR,$$
which is exactly the relation obtained from similarity.
5. Conclude equality of the two sides of $\triangle PQR$
- From the proportion $\frac{PT}{PQ}=\frac{PS}{PR}$ and the equality $PS\cdot PQ = PT\cdot PR$, we deduce
$$PQ = PR.$$
- Therefore, the sides $PQ$ and $PR$ of $\triangle PQR$ are equal.
6. Result
- Since two sides of $\triangle PQR$ are equal, $\triangle PQR$ is an isosceles triangle (with $PQ = PR$).
Question 3
Hint available
In Fig. 6.29, if PQ || RS, prove that POQ ~ SOR. Fig. 6.29
Key Idea
When a pair of corresponding sides of two triangles are parallel, the alternate interior angles formed with the transversals are equal. Hence two angles of one triangle are equal to two angles of the other triangle, establishing similarity by the AA (Angle‑Angle) criterion.
Step-by-Step Solution
1. Identify the parallel sides\
Given PQ \parallel RS.\
\
2. Use alternate interior angles\
- The line \(QO\) cuts the two parallel lines PQ and RS. Therefore\
$$\angle PQO = \angle RSO \quad\text{(alternate interior angles)}.$$\
- The line \(PO\) also cuts the two parallel lines PQ and RS. Hence\
$$\angle QPO = \angle SRO \quad\text{(alternate interior angles)}.$$\
\
3. Third angle equality\
In any triangle, the sum of the interior angles is \(180^{\circ}\).\
For \(\triangle POQ\):\
$$\angle POQ = 180^{\circ} - (\angle PQO + \angle QPO).$$\
For \(\triangle SOR\):\
$$\angle SOR = 180^{\circ} - (\angle RSO + \angle SRO).$$\
Since \(\angle PQO = \angle RSO\) and \(\angle QPO = \angle SRO\), the remaining angles are also equal, i.e.,\
$$\angle POQ = \angle SOR.$$\
\
4. Apply AA similarity criterion\
We have shown that two angles of \(\triangle POQ\) are respectively equal to two angles of \(\triangle SOR\):\
$$\angle PQO = \angle RSO, \quad \angle QPO = \angle SRO.$$\
Hence, by the AA criterion,\
$$\triangle POQ \sim \triangle SOR.$$\
\
5. Conclusion\
Therefore, when PQ is parallel to RS, the triangles POQ and SOR are similar.
Given PQ \parallel RS.\
\
2. Use alternate interior angles\
- The line \(QO\) cuts the two parallel lines PQ and RS. Therefore\
$$\angle PQO = \angle RSO \quad\text{(alternate interior angles)}.$$\
- The line \(PO\) also cuts the two parallel lines PQ and RS. Hence\
$$\angle QPO = \angle SRO \quad\text{(alternate interior angles)}.$$\
\
3. Third angle equality\
In any triangle, the sum of the interior angles is \(180^{\circ}\).\
For \(\triangle POQ\):\
$$\angle POQ = 180^{\circ} - (\angle PQO + \angle QPO).$$\
For \(\triangle SOR\):\
$$\angle SOR = 180^{\circ} - (\angle RSO + \angle SRO).$$\
Since \(\angle PQO = \angle RSO\) and \(\angle QPO = \angle SRO\), the remaining angles are also equal, i.e.,\
$$\angle POQ = \angle SOR.$$\
\
4. Apply AA similarity criterion\
We have shown that two angles of \(\triangle POQ\) are respectively equal to two angles of \(\triangle SOR\):\
$$\angle PQO = \angle RSO, \quad \angle QPO = \angle SRO.$$\
Hence, by the AA criterion,\
$$\triangle POQ \sim \triangle SOR.$$\
\
5. Conclusion\
Therefore, when PQ is parallel to RS, the triangles POQ and SOR are similar.
Question 4
Hint available
Observe Fig. 6.30 and then find P. Fig. 6.30
Key Idea
Use the Exterior Angle Theorem: the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
Step-by-Step Solution
1. Identify the triangle in Fig. 6.30. Let the triangle be \(\triangle ABC\) with \(\angle A = 30^{\circ}\) and \(\angle C = 40^{\circ}\).\
2. The angle \(\angle P\) is the exterior angle at vertex \(B\) (i.e., the angle formed by extending side \(BC\) to point \(P\)).\
3. According to the Exterior Angle Theorem,\
$$\angle P = \angle A + \angle C.$$\
4. Substitute the known interior angles:\
$$\angle P = 30^{\circ} + 40^{\circ} = 70^{\circ}.$$\
5. Hence, the required angle \(\angle P\) measures \(70^{\circ}\).
2. The angle \(\angle P\) is the exterior angle at vertex \(B\) (i.e., the angle formed by extending side \(BC\) to point \(P\)).\
3. According to the Exterior Angle Theorem,\
$$\angle P = \angle A + \angle C.$$\
4. Substitute the known interior angles:\
$$\angle P = 30^{\circ} + 40^{\circ} = 70^{\circ}.$$\
5. Hence, the required angle \(\angle P\) measures \(70^{\circ}\).
Question 5
Hint available
In Fig. 6.31, OA . OB = OC . OD. Show that A = C and B = D.
Key Idea
The given relation OA·OB = OC·OD is the condition for the four points A, B, C, D to be concyclic (Power of a Point theorem). Once the quadrilateral ABCD is cyclic, equal angles subtend equal chords: ∠A and ∠C subtend the same chord BD, and ∠B and ∠D subtend the same chord AC.
Step-by-Step Solution
1. Given: OA·OB = OC·OD, where O is an interior point and OA, OB, OC, OD are straight lines meeting the circle at A, B, C, D respectively.\
2. Power of a Point: For a point O inside a circle, if two chords AB and CD intersect at O, then OA·OB = OC·OD. The converse is also true – if OA·OB = OC·OD, the four points A, B, C, D lie on a single circle. Hence, A, B, C, D are concyclic.\
3. Cyclic Quadrilateral Property: In a cyclic quadrilateral, equal chords subtend equal angles at the circumference.\
4. Identify the chords: In the cyclic quadrilateral ABCD, chord BD subtends ∠A at point A and ∠C at point C. Similarly, chord AC subtends ∠B at point B and ∠D at point D.\
5. Apply the property: Since the same chord BD subtends ∠A and ∠C, we have \(\angle A = \angle C\). Likewise, the same chord AC subtends \(\angle B\) and \(\angle D\), giving \(\angle B = \angle D\).\
6. Conclusion: Hence, \(\angle A = \angle C\) and \(\angle B = \angle D\).
2. Power of a Point: For a point O inside a circle, if two chords AB and CD intersect at O, then OA·OB = OC·OD. The converse is also true – if OA·OB = OC·OD, the four points A, B, C, D lie on a single circle. Hence, A, B, C, D are concyclic.\
3. Cyclic Quadrilateral Property: In a cyclic quadrilateral, equal chords subtend equal angles at the circumference.\
4. Identify the chords: In the cyclic quadrilateral ABCD, chord BD subtends ∠A at point A and ∠C at point C. Similarly, chord AC subtends ∠B at point B and ∠D at point D.\
5. Apply the property: Since the same chord BD subtends ∠A and ∠C, we have \(\angle A = \angle C\). Likewise, the same chord AC subtends \(\angle B\) and \(\angle D\), giving \(\angle B = \angle D\).\
6. Conclusion: Hence, \(\angle A = \angle C\) and \(\angle B = \angle D\).
Question 6
Hint available
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Key Idea
The lamp, the top of the girl and the tip of the shadow form two similar right‑angled triangles. By using the property of similar triangles, the ratio of corresponding sides gives a relation between the distance of the girl from the lamp and the length of her shadow.
Step-by-Step Solution
1. Convert all quantities to the same unit\
Height of girl = 90 cm = 0.9 m.\
Height of lamp = 3.6 m (already in metres).\
2. Find the distance of the girl from the lamp after 4 s\
Speed = 1.2 m/s, time = 4 s\
\[ d = vt = 1.2 \times 4 = 4.8\ \text{m} \]\
So the girl is 4.8 m away from the base of the lamp.
3. Set up the similar‑triangle relation\
Let \(x\) be the length of the shadow (in metres).\
\[\frac{\text{Height of lamp}}{\text{Distance from lamp to tip of shadow}} = \frac{\text{Height of girl}}{\text{Length of shadow}}\]\
The distance from the lamp to the tip of the shadow = \(d + x = 4.8 + x\).\
Hence,\
\[ \frac{3.6}{4.8 + x} = \frac{0.9}{x} \]
4. Solve for \(x\)\
Cross‑multiply: \(3.6x = 0.9(4.8 + x)\).\
Expand: \(3.6x = 4.32 + 0.9x\).\
Bring like terms together: \(3.6x - 0.9x = 4.32\) → \(2.7x = 4.32\).\
\[ x = \frac{4.32}{2.7} = 1.6 \ \text{m} \]
5. State the answer\
The length of the girl's shadow after 4 seconds is 1.6 metres.
Height of girl = 90 cm = 0.9 m.\
Height of lamp = 3.6 m (already in metres).\
2. Find the distance of the girl from the lamp after 4 s\
Speed = 1.2 m/s, time = 4 s\
\[ d = vt = 1.2 \times 4 = 4.8\ \text{m} \]\
So the girl is 4.8 m away from the base of the lamp.
3. Set up the similar‑triangle relation\
Let \(x\) be the length of the shadow (in metres).\
\[\frac{\text{Height of lamp}}{\text{Distance from lamp to tip of shadow}} = \frac{\text{Height of girl}}{\text{Length of shadow}}\]\
The distance from the lamp to the tip of the shadow = \(d + x = 4.8 + x\).\
Hence,\
\[ \frac{3.6}{4.8 + x} = \frac{0.9}{x} \]
4. Solve for \(x\)\
Cross‑multiply: \(3.6x = 0.9(4.8 + x)\).\
Expand: \(3.6x = 4.32 + 0.9x\).\
Bring like terms together: \(3.6x - 0.9x = 4.32\) → \(2.7x = 4.32\).\
\[ x = \frac{4.32}{2.7} = 1.6 \ \text{m} \]
5. State the answer\
The length of the girl's shadow after 4 seconds is 1.6 metres.
Question 7
Hint available
In Fig. 6.33, CM and RN are respectively the medians of ABC and PQR. If ABC ~ PQR, prove that : (i) AMC ~ PNR (ii) CM AB RN PQ (iii) CMB ~ RNQ
Key Idea
In similar triangles, the ratios of corresponding sides are equal. The midpoint of a side divides the side into two equal parts, therefore a median of a triangle is a line joining a vertex to the midpoint of the opposite side. Consequently, the median of one triangle corresponds to the median of the similar triangle, and the triangles formed by a vertex, the midpoint of the opposite side and the third vertex are also similar.
Step-by-Step Solution
1. Given data
* \(\Delta ABC \sim \Delta PQR\). Hence
$$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=k \quad (k>0).$$
* \(M\) is the midpoint of \(AB\) ⇒ \(AM = MB = \dfrac{AB}{2}\).
* \(N\) is the midpoint of \(PQ\) ⇒ \(PN = NQ = \dfrac{PQ}{2}\).
* \(CM\) and \(RN\) are medians.
2. Proof of (i) \(\Delta AMC \sim \Delta PNR\)
* In \(\Delta AMC\) the sides are \(AM, AC, CM\).
* In \(\Delta PNR\) the sides are \(PN, PR, RN\).
* Using the midpoint relations:
$$\frac{AM}{PN}=\frac{\dfrac{AB}{2}}{\dfrac{PQ}{2}}=\frac{AB}{PQ}=k.$$
* From similarity of the original triangles:
$$\frac{AC}{PR}=k \quad\text{and}\quad \frac{CM}{RN}=k$$
(the last equality follows because a median is a line joining a vertex to the midpoint of the opposite side; corresponding medians are in the same ratio as the corresponding sides).
* Hence the three pairs of corresponding sides of \(\Delta AMC\) and \(\Delta PNR\) are in the same ratio \(k\). By the S.S.S. criterion, the two triangles are similar.
3. Proof of (ii) \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\)
* From step‑2 we already have \(\dfrac{CM}{RN}=\dfrac{AB}{PQ}\).
* Cross‑multiplying gives \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\).
* Thus the ratio of a median to its corresponding side is the same in both triangles.
4. Proof of (iii) \(\Delta CMB \sim \Delta RNQ\)
* The sides of \(\Delta CMB\) are \(CM, MB, CB\).
* The sides of \(\Delta RNQ\) are \(RN, NQ, RQ\).
* Using the midpoint relations and the similarity of the original triangles:
\[\frac{CM}{RN}=\frac{AB}{PQ}=k,\qquad \frac{MB}{NQ}=\frac{\dfrac{AB}{2}}{\dfrac{PQ}{2}}=\frac{AB}{PQ}=k,\qquad \frac{CB}{RQ}=\frac{BC}{QR}=k.\]
* All three pairs of corresponding sides are in the same ratio \(k\); therefore, by the S.S.S. criterion, \(\Delta CMB\) is similar to \(\Delta RNQ\).
5. Conclusion
* Hence, all three required statements are proved.
* The key steps rely on (i) the definition of a median, (ii) the property that corresponding medians of similar triangles are in the same ratio as the corresponding sides, and (iii) the S.S.S. similarity criterion.
* \(\Delta ABC \sim \Delta PQR\). Hence
$$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=k \quad (k>0).$$
* \(M\) is the midpoint of \(AB\) ⇒ \(AM = MB = \dfrac{AB}{2}\).
* \(N\) is the midpoint of \(PQ\) ⇒ \(PN = NQ = \dfrac{PQ}{2}\).
* \(CM\) and \(RN\) are medians.
2. Proof of (i) \(\Delta AMC \sim \Delta PNR\)
* In \(\Delta AMC\) the sides are \(AM, AC, CM\).
* In \(\Delta PNR\) the sides are \(PN, PR, RN\).
* Using the midpoint relations:
$$\frac{AM}{PN}=\frac{\dfrac{AB}{2}}{\dfrac{PQ}{2}}=\frac{AB}{PQ}=k.$$
* From similarity of the original triangles:
$$\frac{AC}{PR}=k \quad\text{and}\quad \frac{CM}{RN}=k$$
(the last equality follows because a median is a line joining a vertex to the midpoint of the opposite side; corresponding medians are in the same ratio as the corresponding sides).
* Hence the three pairs of corresponding sides of \(\Delta AMC\) and \(\Delta PNR\) are in the same ratio \(k\). By the S.S.S. criterion, the two triangles are similar.
3. Proof of (ii) \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\)
* From step‑2 we already have \(\dfrac{CM}{RN}=\dfrac{AB}{PQ}\).
* Cross‑multiplying gives \(\dfrac{CM}{AB}=\dfrac{RN}{PQ}\).
* Thus the ratio of a median to its corresponding side is the same in both triangles.
4. Proof of (iii) \(\Delta CMB \sim \Delta RNQ\)
* The sides of \(\Delta CMB\) are \(CM, MB, CB\).
* The sides of \(\Delta RNQ\) are \(RN, NQ, RQ\).
* Using the midpoint relations and the similarity of the original triangles:
\[\frac{CM}{RN}=\frac{AB}{PQ}=k,\qquad \frac{MB}{NQ}=\frac{\dfrac{AB}{2}}{\dfrac{PQ}{2}}=\frac{AB}{PQ}=k,\qquad \frac{CB}{RQ}=\frac{BC}{QR}=k.\]
* All three pairs of corresponding sides are in the same ratio \(k\); therefore, by the S.S.S. criterion, \(\Delta CMB\) is similar to \(\Delta RNQ\).
5. Conclusion
* Hence, all three required statements are proved.
* The key steps rely on (i) the definition of a median, (ii) the property that corresponding medians of similar triangles are in the same ratio as the corresponding sides, and (iii) the S.S.S. similarity criterion.