Find the best tutors and institutes for Class 10 Tuition
Q9:
$ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
Given: A trapezium $ABCD$ where $AB \parallel DC$. The diagonals $AC$ and $BD$ intersect each other at point $O$.
To Prove: $\frac{AO}{BO} = \frac{CO}{DO}$
Construction: Draw a line $EO$ through point $O$ such that $EO \parallel AB$. Since $AB \parallel DC$ (given) and $EO \parallel AB$ (by construction), it follows that $EO \parallel DC$.
Step 1: Applying Thales Theorem (Basic Proportionality Theorem) in $\triangle ADC$
In $\triangle ADC$, since $EO \parallel DC$, by the Basic Proportionality Theorem (BPT), the line drawn parallel to one side of a triangle intersecting the other two sides divides the two sides in the same ratio.
Therefore: $\frac{AE}{ED} = \frac{AO}{OC}$ --- (Equation 1) [By BPT]
Step 2: Applying Thales Theorem in $\triangle ABD$
In $\triangle ABD$, since $EO \parallel AB$, by the Basic Proportionality Theorem:
$\frac{DE}{EA} = \frac{DO}{OB}$
Taking the reciprocal of both sides:
$\frac{AE}{ED} = \frac{BO}{DO}$ --- (Equation 2) [By BPT]
Step 3: Equating the results
From Equation 1 and Equation 2, since both expressions are equal to $\frac{AE}{ED}$, we can equate them:
$\frac{AO}{OC} = \frac{BO}{DO}$
Step 4: Rearranging the terms
To obtain the required form $\frac{AO}{BO} = \frac{CO}{DO}$, we perform cross-multiplication or swap the means:
$\frac{AO}{BO} = \frac{CO}{DO}$
Conclusion: By applying the Basic Proportionality Theorem to the triangles formed by the construction of a parallel line, we have successfully shown the required ratio.
Final Answer: $\frac{AO}{BO} = \frac{CO}{DO}$
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).
- 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.
- 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 $\frac{AM}{AB} = \frac{AN}{AD}$.
- Q4: In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.
- 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).
