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Q4:
In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.
In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.
Solution :
Given: In $\triangle ABC$, $DE \parallel AC$ and $DF \parallel AE$.
To Prove: $\frac{BF}{FE} = \frac{BE}{EC}$
Visual Representation:
Theorem Used: Basic Proportionality Theorem (BPT) or Thales Theorem. It states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Step 1: Applying BPT in $\triangle ABC$
In $\triangle ABC$, we are given that $DE \parallel AC$.
According to the Basic Proportionality Theorem, since $DE \parallel AC$, the line $DE$ divides the sides $BA$ and $BC$ proportionally:
$\frac{BD}{DA} = \frac{BE}{EC}$ --- (Equation 1)
[Justification: By Basic Proportionality Theorem, as $DE \parallel AC$]
Step 2: Applying BPT in $\triangle ABE$
In $\triangle ABE$, we are given that $DF \parallel AE$.
According to the Basic Proportionality Theorem, since $DF \parallel AE$, the line $DF$ divides the sides $BA$ and $BE$ proportionally:
$\frac{BD}{DA} = \frac{BF}{FE}$ --- (Equation 2)
[Justification: By Basic Proportionality Theorem, as $DF \parallel AE$]
Step 3: Comparing the Equations
We now have two equations:
From Equation 1: $\frac{BD}{DA} = \frac{BE}{EC}$
From Equation 2: $\frac{BD}{DA} = \frac{BF}{FE}$
Since both expressions are equal to the same ratio $\frac{BD}{DA}$, we can equate them by the Euclid's axiom which states that things which are equal to the same thing are equal to one another.
Therefore, $\frac{BF}{FE} = \frac{BE}{EC}$
Conclusion:
By comparing the ratios derived from the two triangles using the Basic Proportionality Theorem, we have successfully demonstrated the required equality.
Final Answer: Hence Proved, $\frac{BF}{FE} = \frac{BE}{EC}$
More Questions from Class 10 Mathematics Triangles EXERCISE 6.2
- Q1(i): In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $EC$ in (i).
- Q1(ii): In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $AD$ in (ii).
- Q2(i): $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (i) $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm
- Q2(ii): $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (ii) $PE = 4$ cm, $QE = 4.5$ cm, $PF = 8$ cm and $RF = 9$ cm
- Q2(iii): $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$ : (iii) $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm and $PF = 0.36$ cm
- Q3: In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.
- Q5: In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.
- Q6: In Fig. 6.21, $A$, $B$ and $C$ are points on $OP$, $OQ$ and $OR$ respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that $BC \parallel QR$.
- Q7: Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
- Q8: Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
- Q9: $ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
- Q10: The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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