2. In Fig. 6.35, , and . Find , and .

Understand the Question Line segment is a straight line, so and form a linear pair with a sum of . In , the sum of interior angles is , allowing us to find . Since , corresponding angles are equal, which gives . Step 1 · Find Since is a straight line, and form a linear pair. Step 2 · Find In , using the angle sum property of a triangle Step 3 · Find Given . Since corresponding angles of similar triangles are equal Answer > Common Mistakes Incorrect Angle Correspondence: When matching angles in similar triangles , vertex corresponds to (so ), not to vertex . Linear Pair Confusion: Mistaking vertically opposite angles or non-adjacent angles instead of recognizing that and lie on the straight line .

Triangles | Exercise 6.3

Question 2

In Fig. 6.35, $\Delta \text{ODC} \sim \Delta \text{OBA}$ , $\angle \text{BOC} = 125^\circ$ and $\angle \text{CDO} = 70^\circ$ . Find $\angle \text{DOC}$ , $\angle \text{DCO}$ and $\angle \text{OAB}$ .

Solution

Understand the Question

Line segment $\text{BD}$ is a straight line, so $\angle \text{DOC}$ and $\angle \text{BOC}$ form a linear pair with a sum of $180^\circ$ .
In $\Delta \text{ODC}$ , the sum of interior angles is $180^\circ$ , allowing us to find $\angle \text{DCO}$ .
Since $\Delta \text{ODC} \sim \Delta \text{OBA}$ , corresponding angles are equal, which gives $\angle \text{OAB} = \angle \text{DCO}$ .

Step 1 · Find $\angle \text{DOC}$

Since $\text{BD}$ is a straight line, $\angle \text{DOC}$ and $\angle \text{BOC}$ form a linear pair. Diagram 1

\begin{aligned} \angle \text{DOC} + \angle \text{BOC} &= 180^\circ \\ \angle \text{DOC} + 125^\circ &= 180^\circ \\ \angle \text{DOC} &= 180^\circ - 125^\circ \\ &= 55^\circ \end{aligned}

Step 2 · Find $\angle \text{DCO}$

In $\Delta \text{ODC}$ , using the angle sum property of a triangle

\begin{aligned} \angle \text{CDO} + \angle \text{DCO} + \angle \text{DOC} &= 180^\circ \\ 70^\circ + \angle \text{DCO} + 55^\circ &= 180^\circ \\ 125^\circ + \angle \text{DCO} &= 180^\circ \\ \angle \text{DCO} &= 180^\circ - 125^\circ \\ &= 55^\circ \end{aligned}

Step 3 · Find $\angle \text{OAB}$

Given $\Delta \text{ODC} \sim \Delta \text{OBA}$ .

Since corresponding angles of similar triangles are equal

$\angle \text{OAB} = \angle \text{DCO}$

$\angle \text{OAB} = 55^\circ$

Answer

$\angle \text{DOC} = 55^\circ, \quad \angle \text{DCO} = 55^\circ, \quad \angle \text{OAB} = 55^\circ$

Common Mistakes

Incorrect Angle Correspondence: When matching angles in similar triangles $\Delta \text{ODC} \sim \Delta \text{OBA}$ , vertex $\text{C}$ corresponds to $\text{A}$ (so $\angle \text{OAB} = \angle \text{OCD}$ ), not to vertex $\text{D}$ .
Linear Pair Confusion: Mistaking vertically opposite angles or non-adjacent angles instead of recognizing that $\angle \text{DOC}$ and $\angle \text{BOC}$ lie on the straight line $\text{BD}$ .

More questions in Exercise 6.3

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

In Fig. 6.35, $\Delta \text{ODC} \sim \Delta \text{OBA}$ , $\angle \text{BOC} = 125^\circ$ and $\angle \text{CDO} = 70^\circ$ . Find $\angle \text{DOC}$ , $\angle \text{DCO}$ and $\angle \text{OAB}$ .

Diagonals $\text{AC}$ and $\text{BD}$ of a trapezium $\text{ABCD}$ with $\text{AB} \parallel \text{DC}$ intersect each other at the point $\text{O}$ . Using a similarity criterion for two triangles, show that $\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}$ .

In Fig. 6.36, $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$ . Show that $\Delta \text{PQS} \sim \Delta \text{TQR}$ .

S and T are points on sides $PR$ and $QR$ of $\Delta \text{PQR}$ such that $\angle \text{P} = \angle \text{RTS}$ . Show that $\Delta \text{RPQ} \sim \Delta \text{RTS}$ .

In Fig. 6.37, if $\Delta \text{ABE} \cong \Delta \text{ACD}$ , show that $\Delta \text{ADE} \sim \Delta \text{ABC}$ .

In Fig. 6.38, altitudes $AD$ and $CE$ of $\Delta \text{ABC}$ intersect each other at the point $P$ . Show that:

(i) $\Delta \text{AEP} \sim \Delta \text{CDP}$

(ii) $\Delta \text{ABD} \sim \Delta \text{CBE}$

(iii) $\Delta \text{AEP} \sim \Delta \text{ADB}$

(iv) $\Delta \text{PDC} \sim \Delta \text{BEC}$

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta \text{ABE} \sim \Delta \text{CFB}$ .

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) $\Delta \text{ABC} \sim \Delta \text{AMP}$

(ii) $\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}$

Q10

CD and GH are respectively the bisectors of $\angle \text{ACB}$ and $\angle \text{EGF}$ such that D and H lie on sides AB and FE of $\Delta \text{ABC}$ and $\Delta \text{EFG}$ respectively. If $\Delta \text{ABC} \sim \Delta \text{FEG}$ , show that:

(i) $\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}$

(ii) $\Delta \text{DCB} \sim \Delta \text{HGE}$

(iii) $\Delta \text{DCA} \sim \Delta \text{HGF}$

Q11

In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$ . If $AD \perp BC$ and $EF \perp AC$ , prove that $\Delta ABD \sim \Delta ECF$ .

Q12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\Delta PQR$ (see Fig. 6.41). Show that $\Delta ABC \sim \Delta PQR$ .

Q13

D is a point on the side $BC$ of a triangle $ABC$ such that $\angle \text{ADC} = \angle \text{BAC}$ . Show that $\text{CA}^2 = \text{CB} \cdot \text{CD}$ .

Q14

Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$ . Show that $\Delta ABC \sim \Delta PQR$ .

Q15

A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.

Q16

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$ , respectively where $\Delta ABC \sim \Delta PQR$ , prove that

$\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}$