NCERT Solutions to Ex 6.2 Class 10 math on Similar triangles

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NCERT Solutions to Ex 6.2 Class 10 math on Results of Similarity of triangles

Similarity of Triangles and Results of a line segmenting two sides of a triangle proportionately: Solution NCERT Ex 6.2

You'll learn results of similarity of triangles in the form of a line segmenting two sides of a triangle proportionately. NCERT Ex 6.2 class 10 math solved.

We'll cover a few important results of similarity of two triangles with proofs,

A line parallel to one side of a triangle segments the other two sides proportionately.
A line segmenting two sides of a triangle proportionately must be parallel to the third side.
Example problem 1: A line parallel to a parallel side of a trapezium segments other two sides proportionately.
Example Problem 2: A line segmenting two sides of a triangle proportionately with alternate angles equal.
NCERT Ex 6.2 Class 10 math on results of similarity of triangles.
Solution to NCERT Ex 6.2 Class 10 math on results of similarity of triangles.

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Consequence of a line parallel to a side of a triangle intersecting the other two sides at two distinct points

Drawing from polygon similarity concepts we can say, two triangles are similar if,

their corresponding angles are equal, and,

their corresponding sides are in same ratio and are proportional.

We'll know more about the criteria of similarity of two triangles in the next session.

Here, we'll just concentrate on a few results that are related to similarity of triangles without explicitly mentioning the triangles to be similar.

The first result follows.

Theorem 6.1.

If a line parallel to any side of a triangle intersects the other two sides at two distinct points, then the segments of the two sides created by this parallel line must be proportional to each other.

Proof of Theorem 6.1.

In the following figure, DE parallel to BC intersects AB and AC at D and E respectively. It is to prove,

$\displaystyle\frac{AD}{BD}=\frac{AE}{CE}$

Proof of Theorem 6.1

To prove the proportionality of the segments from very basic concepts without using the properties of similar triangles, we will take help of ratios of areas of $\triangle ADE$ to $\triangle BDE$ and $\triangle ADE$ to $\triangle CED$ and then we will use the equality of the areas of two triangles $\triangle BDE$ and $\triangle CED$.

The height of $\triangle ADE$ with base as $AD$ is $EM$ which is also the height of $\triangle BDE$ with $BD$ as base. When you take the ratio of areas of these two triangles, the height will cancel out leaving the ratio of $AD$ to $BD$.

Area of $\triangle ADE$ with base as $AD$ and height as $EM$ is,

$\text{Area of }\triangle ADE=\frac{1}{2}AD.EM$.

The same $EM$ is the height of $\triangle BDE$ with BD as base. So,

$\text{Area of }\triangle BDE=\frac{1}{2}BD.EM$.

Ratio of these two areas,

$\displaystyle\frac{\text{Area of }\triangle ADE}{\text{Area of }\triangle BDE}=\frac{AD}{BD}$ $\qquad \qquad \qquad \text{......(1)}$

Let us turn our attention to the right part of $\triangle ABC$.

The height of $\triangle ADE$ with base as $AE$ is $DN$ which is also the height of $\triangle CED$ with $CE$ as base. When you take the ratio of areas of these two triangles, the height will cancel out leaving the ratio of $AE$ to $CE$.

Area of $\triangle ADE$ with base as $AE$ and height as $DN$ is,

$\text{Area of }\triangle ADE=\frac{1}{2}AE.DN$.

The same $DN$ is the height of $\triangle CED$ with $CE$ as base. So,

$\text{Area of }\triangle CED=\frac{1}{2}CE.DN$.

Ratio of these two areas,

$\displaystyle\frac{\text{Area of }\triangle ADE}{\text{Area of }\triangle CED}=\frac{AE}{CE}$ $\qquad \qquad \qquad \text{......(2)}$

In the LHSs of equation 1 and equation 2, identify the key pattern that taking bases as same $DE$, height $h_1$ of $\triangle BDE$ is equal to height $h_2$ of $\triangle CED$.

This is so because $DE \parallel BC$.

It follows, $\text{Area of }\triangle BDE=\text{Area of }\triangle CED$.

As LHSs of equation 1 and equation 2 are equal, RHSs also are equal,

$\displaystyle\frac{AD}{BD}=\frac{AE}{CE}$ Proved.

Now we'll see the result just opposite to Theorem 6.1.

Consequence of a line segmenting two sides of a triangle proportionately

Theorem 6.2.

If a straight line segments two sides of a triangle proportionately, then it must be parallel to the third side of the triangle.

Proof of Theorem 6.2.

In the proof, we'll use only basic concepts with nearly exactly same steps as the above proof except for the last few steps.

In the following figure, line DE segments two sides of $\triangle ABC$ in equal proportions,

$\displaystyle\frac{AD}{BD}=\frac{AE}{CE}$.

To prove, $DE \parallel BC$.

Ratio of areas of two triangles,

$\displaystyle\frac{\text{Area of }\triangle ADE}{\text{Area of }\triangle BDE}=\frac{AD}{BD}$.

Ratio of areas of other pair of triangles,

$\displaystyle\frac{\text{Area of }\triangle ADE}{\text{Area of }\triangle CED}=\frac{AE}{CE}$.

As, $\displaystyle\frac{AD}{BD}=\frac{AE}{CE}$ the two denominators in two LHSs must be equal,

$\text{Area of }\triangle BDE=\text{Area of }\triangle CED$.

With bases as common $DE$,

$\text{Area of }\triangle BDE =\frac{1}{2}DE.h_1$, and,

$\text{Area of }\triangle CED =\frac{1}{2}DE.h_2$.

The two areas being equal,

$h_1=h_2$.

The equal length $h_1$ and $h_2$ being the shortest distance from point $B$ to straight line $DE$ and point $C$ to same straight line $DE$ respectively, these are perpendicular to line $DE$ and parallel to each other.

Finally, $B$ and $C$ being on the same straight line $BC$, quadrangle $BFGC$ must a rectangle and,

$DE \parallel BC$. Proved.

Example Problem 1: A line parallel to one of the two parallel sides of a trapezium segments the two non-parallel sides proportionately

In the following figure, the line EG parallel to one of the two parallel sides BC of the trapezium ABCD intersects the two non-parallel sides at E and G.

Theorem 6.1 is also applicable in this case.

EG will segment AB and DC in proportional segments,

$\displaystyle\frac{AE}{BE}=\frac{DG}{CG}$.

Proof of application of Theorem 6.1 in a trapezium

The diagonal AC divides the trapezium into two triangles, each with two sides intersected by the same line EG parallel to their bases.

Applying Theorem 6.1 on the $\triangle ABC$, as $EF \parallel BC$,

$\displaystyle\frac{AE}{BE}=\frac{AF}{CF}$.

Similarly applying Theorem 6.1 on $\triangle CDA$, as $GF \parallel AD$,

$\displaystyle\frac{AF}{CF}=\frac{DG}{CG}$.

Joining the two equations by the common link equality term $\displaystyle\frac{AF}{CF}$,

$\displaystyle\frac{AE}{BE}=\frac{DG}{CG}$.

Proved.

Example Problem 2: A line segmenting two sides of a triangle proportionately with alternate angles equal

In the following figure, the line DE segments the two sides of the $\triangle ABC$ proportionately,

$\displaystyle\frac{AD}{BD}=\frac{AE}{CE}$.

Also $\angle DEA=\angle ABC$.

Show that the $\triangle ABC$ is isosceles.

Proof of Example problem 2

Applying Theorem 6.2 on the triangle system, as DE segments AB and AC proportionately,

$DE \parallel BC$.

So, pair of external angles are equal,

$\angle ADE=\angle ABC=\angle DEA=\angle ACB$.

$\triangle ABC$ is isosceles with its two base angles equal.

Proved.

NCERT Ex 6.2 Class 10 math on results of similarity of triangles

Problem 1.

In figure below, in (i) and (ii) $DE \parallel BC$. Find EC in (i) and AD and in (ii).

Problem 2.

E and F are points on the sides PQ and PR respectively in a $\triangle PQR$. For each of the following cases state whether $EF \parallel QR$.

PE=3.9 cm, EQ=3 cm, PF=3.6 cm and FR=2.4 cm.
PE=4 cm, EQ=4.5 cm, PF=8 cm and FR=9 cm.
PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm and PF=0.36 cm.

Problem 3.

In figure below, if $LM \parallel CB$ and $LN \parallel CD$ prove that $\displaystyle\frac{AM}{AB}=\frac{AN}{AD}$.

Problem 4.

In figure below, $DE \parallel AC$ and $DF \parallel AE$. Prove that, $\displaystyle\frac{BF}{FE}=\frac{BE}{EC}$.

Problem 5.

In Figure below, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.

Problem 6.

In figure below, A, B and C are points on OP, OQ and OR respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that, $BC \parallel QR$.

Problem 7.

Using Theorem 6.1 prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Problem 8.

Using Theorem 6.2 prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Problem 9.

ABCD is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at point O. Show that $\displaystyle\frac{AO}{BO}=\frac{CO}{DO}$.

Problem 10.

The diagonals of a quadrilateral ABCD intersect each other at point O such that $\displaystyle\frac{AO}{BO}=\frac{CO}{DO}$. Show that ABCD is a trapezium.

Solution to NCERT Ex 6.2 Class 10 math on results of similarity of triangles

Problem 1.

In figure below, in (i) and (ii) $DE \parallel BC$. Find EC in (i) and AD and in (ii).

Solution to problem 1.

i. The segment ratios of two sides AB and AC in figure (i) are,

$\displaystyle\frac{AD}{DB}=\frac{1.5}{3}=0.5$, and

$\displaystyle\frac{AE}{EC}=\frac{1}{EC}$.

With $DE \parallel BC$, by Theorem 6.1, the two segment ratios are equal,

$\displaystyle\frac{1}{EC}=0.5$, Or, $EC=2$ cm.

Answer: $EC=2$ cm.

ii. The segment ratios of two sides AB and AC in figure (ii) are,

$\displaystyle\frac{AD}{DB}=\frac{AD}{7.2}$, and,

$\displaystyle\frac{AE}{EC}=\frac{1.8}{5.4}$.

With $DE \parallel BC$, by Theorem 6.1, the two segment ratios are equal,

$\displaystyle\frac{AD}{7.2}=\frac{1.8}{5.4}=\frac{1}{3}$,

Or, $AD=2.4$ cm.

Answer: $AD=2.4$ cm.

Problem 2.i.

E and F are points on the sides PQ and PR respectively in a $\triangle PQR$. For the following case state whether $EF \parallel QR$.

i. PE=3.9 cm, EQ=3 cm, PF=3.6 cm and FR=2.4 cm.

Solution to problem 2.i.

In the above figure, segment ratio for PQ is,

$\displaystyle\frac{PE}{EQ}=\frac{3.9}{3}=3$,

And segment ratio for PR is,

$\displaystyle\frac{PF}{FR}=\frac{3.6}{2.4}=1.5$.

Ratios of segments on sides PQ and PR made by intersecting line EF being unequal, and segments not proportionate, by Theorem 6.2, EF is not parallel to QR.

Answer: No.

Problem 2.ii.

E and F are points on the sides PQ and PR respectively in a $\triangle PQR$. For the following case state whether $EF \parallel QR$.

ii. PE=4 cm, EQ=4.5 cm, PF=8 cm and FR=9 cm.

Solution to problem 2.ii.

Segment ratios of sides PQ and PR made at E and F are respectively,

$\displaystyle\frac{EQ}{PE}=\frac{4.5}{4}=1.125$, and

$\displaystyle\frac{FR}{PF}=\frac{9}{8}=1.125$.

The two segment ratios being equal, the two sides are proportionately segmented by line EF.

By Theorem 6.2 $EF \parallel QR$.

Answer: Yes. $EF \parallel QR$.

Problem 2.iii.

E and F are points on the sides PQ and PR respectively in a $\triangle PQR$. For the following case state whether $EF \parallel QR$.

iii. PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm and PF=0.36 cm.

Solution to problem 2.iii.

Segment $EQ=PQ-PE=1.28-0.18=1.1$ cm, and,

Segment $FR=PR-PF=2.56-0.36=2.2$ cm.

The segment ratios are,

$\displaystyle\frac{PE}{EQ}=\frac{0.18}{1.1}$, and,

$\displaystyle\frac{PF}{FR}=\frac{0.36}{2.2}=\frac{0.18}{1.1}=\displaystyle\frac{PE}{EQ}$,

$\Rightarrow$ EF segments two sides PQ and PR proportionately,

$\Rightarrow$ By Theorem 6.2, $EF \parallel QR$.

Answer: Yes. $EF \parallel QR$.

Problem 3.

In figure below, if $LM \parallel CB$ and $LN \parallel CD$ prove that $\displaystyle\frac{AM}{AB}=\frac{AN}{AD}$.

Solution to problem 3.

Identify the key pattern, side AC is common between two pairs of sides segmented, AB, AC and AC, AD.

In $\triangle ABC$ $LM \parallel BC$.

By Theorem 6.1 LM segments the two sides AB and AC proportionately,

$\displaystyle\frac{BM}{AM}=\frac{CL}{AL}$ $\qquad \qquad \qquad ......(1)$

In $\triangle ADC$ $LN \parallel CD$.

By Theorem 6.1 LN segments the two sides AC and AD proportionately,

$\displaystyle\frac{CL}{AL}=\frac{DN}{AN}$ $\qquad \qquad \qquad ......(2)$

From equations (1) and (2),

$\displaystyle\frac{BM}{AM}=\frac{DN}{AN}$.

Add 1 to both sides of the equation,

$\displaystyle\frac{AB}{AM}=\frac{AD}{AN}$, as $(BM+AM)=AB$, and $(DN+AN)=AD$.

Inverting,

$\displaystyle\frac{AM}{AB}=\frac{AN}{AD}$.

Proved.

Problem 4.

In figure below, $DE \parallel AC$ and $DF \parallel AE$. Prove that, $\displaystyle\frac{BF}{FE}=\frac{BE}{EC}$.

Solution to problem 4.

Sides AB, BC and AB, BE are segmented proportionately by DE and DF respectively, where segments on AB are same for both cases.

This is the key pattern identification, and from the two resulting proportionality equations, we'll use the ratio $\displaystyle\frac{BD}{AD}$ as the link equality term to derive the desired the segment equality relation.

With strategy clearly defined, we take up $\triangle ABC$ in which $DE \parallel AC$.

By Theorem 6.1 DE segments AB and BC proportionately,

$\displaystyle\frac{BD}{AD}=\frac{BE}{EC}$ $\qquad \qquad \qquad ......(1)$

In $\triangle ABE$, $DF \parallel AE$.

By Theorem 6.1 DF segments sides AB and BE proportionately,

$\displaystyle\frac{BD}{AD}=\frac{BF}{FE}$ $\qquad \qquad \qquad ......(2)$

From equation (1) and (2) through the link equality term $\displaystyle\frac{BD}{AD}$,

$\displaystyle\frac{BF}{FE}=\frac{BE}{EC}$.

Proved.

Problem 5.

In Figure below, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.

Solution to problem 5.

Key pattern identification and strategy to be followed:

Sides PQ, PO are segmented proportionately by DE and sides PO and PR by DF. The common segments on side PO will be used for getting the equal proportionate segments on PQ and PR.

In $\triangle AQO$, $DE \parallel OQ$ segments sides PQ and PO proportionately.

By Theorem 6.1,

$\displaystyle\frac{PE}{EQ}=\frac{PD}{OD}$ $\qquad \qquad \qquad ......(1)$

In $\triangle ARO$, $DF \parallel OR$ segments sides AO and PR proportionately.

By Theorem 6.1,

$\displaystyle\frac{PD}{OD}=\frac{PF}{FR}$ $\qquad \qquad \qquad ......(2)$

From equation (1) and (2) using the link equality term $\displaystyle\frac{PD}{OD}$,

$\displaystyle\frac{PE}{EQ}=\frac{PF}{FR}$,

$\Rightarrow$ EF segments the two sides PQ and PR of $\triangle PQR$ proportionately,

$\Rightarrow$ By Theorem 6.2, $EF \parallel QR$.

Proved.

Problem 6.

In figure below, A, B and C are points on OP, OQ and OR respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that, $BC \parallel QR$.

Solution to problem 6.

In $\triangle PQO$, PO, QO are segmented by line $AB \parallel PQ$.

By Theorem 6.1 sides PO and QO are segmented by AB proportionately,

$\displaystyle\frac{OA}{AP}=\frac{OB}{BQ}$ $\qquad \qquad \qquad ......(1)$

In $\triangle PRO$, PO, RO are segmented by line $AC \parallel PR$.

By Theorem 6.1 sides PO and RO are segmented proportionately,

$\displaystyle\frac{OA}{AP}=\frac{OC}{CR}$ $\qquad \qquad \qquad ......(2)$

From equation (1) and (2),

$\displaystyle\frac{OB}{BQ}=\frac{OC}{CR}$.

It means, in $\triangle OQR$, line BC segments two sides OQ and OR proportionately.

By Theorem 6.2, $BC \parallel QR$.

Proved.

Problem 7.

Using Theorem 6.1 prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution to problem 7.

Assume line DE parallel to BC of $\triangle ABC$ bisects side AB at D and intersects side AC at E.

By Theorem 6.1 as $DE \parallel BC$, it segments AB and AC proportionately, that is,

$\displaystyle\frac{AD}{DB}=\frac{AE}{EC}$,

$\Rightarrow \displaystyle\frac{AE}{EC}=1$, as $AD=DB$.

$\Rightarrow AE=EC$.

That is, DE bisects side AC also.

Proved.

Problem 8.

Using Theorem 6.2 prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution to problem 8.

Assume DE joins the mid-points D and E of two sides AB and AC of $\triangle ABC$.

D and E being the mid-points of AB and AC,

$AD=DB$, Or, $\displaystyle\frac{AD}{DB}=1$, and,

$AE=EC$, Or, $\displaystyle\frac{AE}{EC}=1$.

The line DE segments two sides AB and AC proportionately.

By Theorem 6.2 then, $DE \parallel BC$.

Proved.

Problem 9.

ABCD is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at point O. Show that $\displaystyle\frac{AO}{BO}=\frac{CO}{DO}$.

Solution to problem 9.

In the following figure representing the problem, the two sides AB and DC are parallel and diagonals AC and BD intersect each other at O. A third line $OF \parallel AB \parallel DC$ is drawn.

Select pair of triangles $\triangle ABC$ and $\triangle BCD$ with bases AB and CD and common side BC.

In $\triangle ABC$, $OF \parallel AB$.

By Theorem 6.1,

$\displaystyle\frac{AO}{CO}=\frac{BF}{FC}$.

Similarly, in $\triangle BCD$, $OF \parallel CD$.

By Theorem 6.1,

$\displaystyle\frac{BO}{DO}=\frac{BF}{FC}$.

Joining the two equations through link equality term $\displaystyle\frac{BF}{FC}$,

$\displaystyle\frac{AO}{CO}=\displaystyle\frac{BO}{DO}$,

$\Rightarrow \displaystyle\frac{AO}{BO}=\displaystyle\frac{CO}{DO}$.

Proved.

Problem 10.

The diagonals of a quadrilateral ABCD intersect each other at point O such that $\displaystyle\frac{AO}{BO}=\frac{CO}{DO}$. Show that ABCD is a trapezium.