D is a point on the side of a triangle such that . Show that .

Understand the Question To show , we can express the relation as a ratio of corresponding sides: . This suggests proving that and are similar. We use the AA (Angle-Angle) similarity criterion by showing that two pairs of corresponding angles are equal ($ = $ common), and then cross-multiply the resulting proportional side lengths. Step 1 · Prove Similarity of Triangles In and : Therefore, by AA similarity criterion: Step 2 · Use Ratio of Corresponding Sides Since corresponding sides of similar triangles are proportional: Taking the second and third ratios: Cross-multiplying: Answer > Hence proved, . Common Mistakes Incorrect Triangle Ordering: Writing instead of . The order of vertices must match equal angles: , , and . Selecting the Wrong Sub-triangle: Attempting to prove similarity for instead of , which does not have the given equal angle .

Triangles | Exercise 6.3

Question 13

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

Solution

Understand the Question

To show $\text{CA}^2 = \text{CB} \cdot \text{CD}$ , we can express the relation as a ratio of corresponding sides: $\dfrac{\text{CD}}{\text{CA}} = \dfrac{\text{CA}}{\text{CB}}$ .
This suggests proving that $\triangle \text{ADC}$ and $\triangle \text{BAC}$ are similar.
We use the AA (Angle-Angle) similarity criterion by showing that two pairs of corresponding angles are equal ( $\angle \text{ADC} = \angle \text{BAC}$ given, and $\angle \text{C}$ common), and then cross-multiply the resulting proportional side lengths.

Step 1 · Prove Similarity of Triangles

In $\Delta \text{ADC}$ and $\Delta \text{BAC}$ : Diagram 1

\begin{aligned} \angle \text{ADC} &= \angle \text{BAC} \quad [\text{Given}] \\[0.6em] \angle \text{ACD} &= \angle \text{BCA} \quad [\text{Common angle}] \end{aligned}

Therefore, by AA similarity criterion: $\Delta \text{ADC} \sim \Delta \text{BAC}$

Step 2 · Use Ratio of Corresponding Sides

Since corresponding sides of similar triangles are proportional: $\dfrac{\text{AD}}{\text{BA}} = \dfrac{\text{DC}}{\text{AC}} = \dfrac{\text{AC}}{\text{BC}}$

Taking the second and third ratios: $\dfrac{\text{DC}}{\text{AC}} = \dfrac{\text{AC}}{\text{BC}}$

Cross-multiplying:

\begin{aligned} \text{DC} \times \text{BC} &= \text{AC} \times \text{AC} \\[0.6em] \text{CD} \cdot \text{CB} &= \text{CA}^2 \end{aligned}

$\text{CA}^2 = \text{CB} \cdot \text{CD}$

Answer

Hence proved, $\text{CA}^2 = \text{CB} \cdot \text{CD}$ .

Common Mistakes

Incorrect Triangle Ordering: Writing $\Delta \text{ADC} \sim \Delta \text{ABC}$ instead of $\Delta \text{ADC} \sim \Delta \text{BAC}$ . The order of vertices must match equal angles: $\text{D} \leftrightarrow \text{A}$ , $\text{C} \leftrightarrow \text{C}$ , and $\text{A} \leftrightarrow \text{B}$ .
Selecting the Wrong Sub-triangle: Attempting to prove similarity for $\Delta \text{ABD}$ instead of $\Delta \text{ADC}$ , which does not have the given equal angle $\angle \text{ADC}$ .

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

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