In Fig. 6.40, is a point on side produced of an isosceles triangle with . If and , prove that .

Understand the Question In , since , the angles opposite to the equal sides are equal: . Both and contain a right angle () formed by the perpendiculars and . By showing two pairs of corresponding angles are equal, we establish similarity using the AA (Angle-Angle) criterion. Step 1 · Equate Base Angles of the Isosceles Triangle Given an isosceles triangle with . Since angles opposite to equal sides in a triangle are equal: Since lies on , and lie on the same line with on , this can be written as: Step 2 · Apply AA Similarity Criterion In and : Therefore, by AA similarity criterion: Answer > Common Mistakes Vertex Correspondence Error: Not maintaining correct vertex order in the similarity statement. is correct because and . Isosceles Property Confusion: Incorrectly assuming instead of . Angles opposite to equal sides and are and respectively.

Triangles | Exercise 6.3

Question 11

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 $\Delta ABD \sim \Delta ECF$ .

Solution

Understand the Question

In $\Delta ABC$ , since $AB = AC$ , the angles opposite to the equal sides are equal: $\angle ABC = \angle ACB$ .
Both $\Delta ABD$ and $\Delta ECF$ contain a right angle ( $90^\circ$ ) formed by the perpendiculars $AD \perp BC$ and $EF \perp AC$ .
By showing two pairs of corresponding angles are equal, we establish similarity using the AA (Angle-Angle) criterion.

Step 1 · Equate Base Angles of the Isosceles Triangle

Given an isosceles triangle $ABC$ with $AB = AC$ . Diagram 1

Since angles opposite to equal sides in a triangle are equal: $\angle ABC = \angle ACB$

Since $D$ lies on $BC$ , and $E, B, C$ lie on the same line with $F$ on $AC$ , this can be written as: $\angle ABD = \angle ECF \quad \dots (1)$

Step 2 · Apply AA Similarity Criterion

In $\Delta ABD$ and $\Delta ECF$ :

\begin{aligned} \angle ADB &= \angle EFC = 90^\circ \quad (\text{Given } AD \perp BC \text{ and } EF \perp AC) \\[0.6em] \angle ABD &= \angle ECF \quad (\text{From } (1)) \end{aligned}

Therefore, by AA similarity criterion: $\Delta ABD \sim \Delta ECF$

Answer

$\Delta ABD \sim \Delta ECF$

Common Mistakes

Vertex Correspondence Error: Not maintaining correct vertex order in the similarity statement. $\Delta ABD \sim \Delta ECF$ is correct because $\angle B = \angle C$ and $\angle D = \angle F$ .
Isosceles Property Confusion: Incorrectly assuming $\angle A = \angle B$ instead of $\angle B = \angle C$ . Angles opposite to equal sides $AB$ and $AC$ are $\angle C$ and $\angle B$ respectively.

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

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