8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that .

Understand the Question To show that , we use the AA (Angle-Angle) similarity criterion, which states that two triangles are similar if two pairs of corresponding angles are equal. In parallelogram : Opposite angles are equal: . Opposite sides are parallel: . With transversal , the alternate interior angles are equal: . Step 1 · Identify Equal Angles Consider parallelogram with side extended to and line segment intersecting at .In and : 1. Opposite angles of a parallelogram are equal: 2. Since , line with transversal : Step 2 · Apply AA Similarity Criterion In and : Therefore, by the Angle-Angle (AA) similarity criterion: Answer > Common Mistakes Incorrect Transversal: Identifying directly without establishing the correct parallel lines and transversal. The clear alternate interior angle pair is using and transversal . Vertex Order in Similarity: Writing instead of . Corresponding vertices must match: , , .

Triangles | Exercise 6.3

Question 8

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

Solution

Understand the Question

To show that $\Delta \text{ABE} \sim \Delta \text{CFB}$ , we use the AA (Angle-Angle) similarity criterion, which states that two triangles are similar if two pairs of corresponding angles are equal.
In parallelogram $\text{ABCD}$ $ABCD$ :
- Opposite angles are equal: $\angle \text{A} = \angle \text{C}$ .
- Opposite sides are parallel: $\text{AD} \parallel \text{BC} \implies \text{AE} \parallel \text{BC}$ .
- With transversal $\text{BE}$ , the alternate interior angles are equal: $\angle \text{AEB} = \angle \text{CBF}$ .

Step 1 · Identify Equal Angles

Consider parallelogram $\text{ABCD}$ with side $\text{AD}$ extended to $\text{E}$ and line segment $\text{BE}$ intersecting $\text{CD}$ at $\text{F}$ .In $\Delta \text{ABE}$ and $\Delta \text{CFB}$ :

Opposite angles of a parallelogram are equal: $\angle \text{A} = \angle \text{C}$
Since $\text{AD} \parallel \text{BC}$ , line $\text{AE} \parallel \text{BC}$ with transversal $\text{BE}$ : $\angle \text{AEB} = \angle \text{CBF} \quad (\text{Alternate interior angles})$

Step 2 · Apply AA Similarity Criterion

In $\Delta \text{ABE}$ and $\Delta \text{CFB}$ : $\angle \text{A} = \angle \text{C}$ $\angle \text{AEB} = \angle \text{CBF}$

Therefore, by the Angle-Angle (AA) similarity criterion: $\Delta \text{ABE} \sim \Delta \text{CFB}$

Answer

$\Delta \text{ABE} \sim \Delta \text{CFB}$

Common Mistakes

Incorrect Transversal: Identifying $\angle \text{ABE} = \angle \text{CFB}$ directly without establishing the correct parallel lines and transversal. The clear alternate interior angle pair is $\angle \text{AEB} = \angle \text{CBF}$ using $\text{AE} \parallel \text{BC}$ and transversal $\text{BE}$ .
Vertex Order in Similarity: Writing $\Delta \text{ABE} \sim \Delta \text{BFC}$ instead of $\Delta \text{ABE} \sim \Delta \text{CFB}$ . Corresponding vertices must match: $\text{A} \leftrightarrow \text{C}$ , $\text{B} \leftrightarrow \text{F}$ , $\text{E} \leftrightarrow \text{B}$ .

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

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