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Q13:
$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
Solution :
Given: A triangle $ABC$ and a point $D$ on the side $BC$ such that $\angle ADC = \angle BAC$.
To Prove: $CA^2 = CB \cdot CD$.
Step 1: Identifying the Triangles to be Compared
To prove the relationship $CA^2 = CB \cdot CD$, we observe that this can be rewritten as the ratio $\frac{CA}{CB} = \frac{CD}{CA}$. This suggests that we should consider the two triangles $\triangle ADC$ and $\triangle BAC$.
Step 2: Establishing Similarity Criteria
We compare $\triangle ADC$ and $\triangle BAC$ based on the following observations:
1. In $\triangle ADC$ and $\triangle BAC$, we are given that $\angle ADC = \angle BAC$. [Given in the problem statement]
2. In $\triangle ADC$ and $\triangle BAC$, the angle $\angle C$ is common to both triangles. That is, $\angle ACD = \angle BCA$. [Common angle]
Step 3: Applying the AA (Angle-Angle) Similarity Criterion
Since two angles of $\triangle ADC$ are equal to two corresponding angles of $\triangle BAC$, by the AA similarity criterion, the triangles are similar.
Therefore, $\triangle ADC \sim \triangle BAC$. [By AA Similarity Criterion]
Step 4: Utilizing the Properties of Similar Triangles
When two triangles are similar, the ratios of their corresponding sides are equal. Based on the correspondence established in Step 3:
$\frac{AD}{BA} = \frac{DC}{AC} = \frac{AC}{BC}$
[Since corresponding sides of similar triangles are proportional]
Step 5: Deriving the Final Equation
We take the relevant parts of the proportionality ratio from Step 4:
$\frac{DC}{AC} = \frac{AC}{BC}$
By performing cross-multiplication:
$AC \cdot AC = DC \cdot BC$
$AC^2 = BC \cdot DC$
Since $AC$ is the same as $CA$ and $BC$ is the same as $CB$, we can rewrite this as:
$CA^2 = CB \cdot CD$
Final Answer: Hence, it is proved that $CA^2 = CB \cdot CD$.
More Questions from Class 10 Mathematics Triangles EXERCISE 6.3
- Q1(i): 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 : (i)
- Q1(ii): 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 : (ii)
- Q1(iii): 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 : (iii)
- Q1(iv): 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 : (iv)
- Q1(v): 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 : (v)
- Q1(vi): 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 : (vi)
- Q10(i): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (i) $\frac{CD}{GH} = \frac{AC}{FG}$
- Q10(ii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (ii) $\triangle DCB \sim \triangle HGE$
- Q10(iii): $CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that: (iii) $\triangle DCA \sim \triangle HGF$
- 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 $\triangle ABD \sim \triangle ECF$.
- Q12: Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.
- 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 $\triangle ABC \sim \triangle PQR$.
- Q15: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
- Q16: If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
- Q2: In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.
- Q3: Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
- Q4: In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.
- Q5: $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.
- Q6: In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.
- Q7(i): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (i) $\triangle AEP \sim \triangle CDP$
- Q7(ii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (ii) $\triangle ABD \sim \triangle CBE$
- Q7(iii): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iii) $\triangle AEP \sim \triangle ADB$
- Q7(iv): In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that: (iv) $\triangle PDC \sim \triangle BEC$
- Q8: $E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.
- Q9(i): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (i) $\triangle ABC \sim \triangle AMP$
- Q9(ii): In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that: (ii) $\frac{CA}{PA} = \frac{BC}{MP}$
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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