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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) $\triangle DCA \sim \triangle HGF$

Solution :

Given:

1. $\triangle ABC \sim \triangle FEG$

2. $CD$ is the bisector of $\angle ACB$, such that $D$ lies on $AB$.

3. $GH$ is the bisector of $\angle EGF$, such that $H$ lies on $FE$.

To Prove:

$\triangle DCA \sim \triangle HGF$

A C B D F E G H

Step 1: Analyze the implications of the similarity of triangles.

Since $\triangle ABC \sim \triangle FEG$, by the definition of similar triangles (corresponding angles are equal):

$\angle A = \angle F$

$\angle B = \angle E$

$\angle ACB = \angle FEG$

Step 2: Utilize the angle bisector property.

We are given that $CD$ bisects $\angle ACB$ and $GH$ bisects $\angle FEG$.

Therefore, $\angle ACD = \frac{1}{2} \angle ACB$

And, $\angle FGH = \frac{1}{2} \angle FEG$

Since $\angle ACB = \angle FEG$ [From Step 1], it follows that:

$\frac{1}{2} \angle ACB = \frac{1}{2} \angle FEG$

$\angle ACD = \angle FGH$

Step 3: Establish similarity between $\triangle DCA$ and $\triangle HGF$.

In $\triangle DCA$ and $\triangle HGF$:

1. $\angle A = \angle F$ [Proved in Step 1]

2. $\angle ACD = \angle FGH$ [Proved in Step 2]

Step 4: Conclusion by AA Similarity Criterion.

By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Therefore, $\triangle DCA \sim \triangle HGF$.

Final Answer: Since $\angle A = \angle F$ and $\angle ACD = \angle FGH$, by the AA similarity criterion, $\triangle DCA \sim \triangle HGF$ is proved.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


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