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

Understand the Question In an isosceles triangle, angles opposite to equal sides are equal (). To prove that , we use the AA (Angle-Angle) similarity criterion by establishing two pairs of corresponding equal angles: 1. The angles from the isosceles triangle property (). 2. The right angles formed by the perpendiculars (). Step 1 · Establish Equal Angles In , since : Rewriting in terms of the required triangles: Given that and : Therefore: Step 2 · Apply AA Similarity Criterion In and : Therefore, by the AA similarity criterion: Answer > (proved) Common Mistakes Mismatched Angle Correspondence: Ensure angle names strictly correspond to the respective vertices when comparing and (i.e., matches , and matches ). Overcomplicating with Side Lengths: Attempting to use SAS or SSS similarity by trying to calculate side ratios instead of directly applying the straightforward AA similarity criterion.

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 an isosceles triangle, angles opposite to equal sides are equal ( $AB = AC \implies \angle B = \angle C$ ).
To prove that $\Delta ABD \sim \Delta ECF$ $Δ A B D \sim Δ E C F$ , we use the AA (Angle-Angle) similarity criterion by establishing two pairs of corresponding equal angles:
1. The angles from the isosceles triangle property ( $\angle ABD = \angle ECF$ ).
2. The right angles formed by the perpendiculars ( $\angle ADB = \angle EFC = 90^\circ$ ).

Step 1 · Establish Equal Angles

Diagram 1

In $\Delta ABC$ , since $AB = AC$ : $\angle ABC = \angle ACB \quad (\text{Angles opposite to equal sides are equal})$

Rewriting in terms of the required triangles: $\angle ABD = \angle ECF \quad \dots (1)$

Given that $AD \perp BC$ and $EF \perp AC$ : $\angle ADB = 90^\circ$ $\angle EFC = 90^\circ$

Therefore: $\angle ADB = \angle EFC = 90^\circ \quad \dots (2)$

Step 2 · Apply AA Similarity Criterion

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

\begin{aligned} \angle ABD &= \angle ECF && [\text{From }(1)] \\[0.6em] \angle ADB &= \angle EFC && [\text{From }(2)] \end{aligned}

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

Answer

$\Delta ABD \sim \Delta ECF$ (proved)

Common Mistakes

Mismatched Angle Correspondence: Ensure angle names strictly correspond to the respective vertices when comparing $\Delta ABD$ and $\Delta ECF$ (i.e., $\angle ABD$ matches $\angle ECF$ , and $\angle ADB$ matches $\angle EFC$ ).
Overcomplicating with Side Lengths: Attempting to use SAS or SSS similarity by trying to calculate side ratios instead of directly applying the straightforward AA similarity criterion.

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

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