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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.
Solution :
Given: A quadrilateral $ABCD$ where the diagonals $AC$ and $BD$ intersect at point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$.
To Prove: Quadrilateral $ABCD$ is a trapezium (i.e., $AB \parallel DC$ or $AD \parallel BC$).
Step 1: Rearranging the given ratio
We are given the equation:
$\frac{AO}{BO} = \frac{CO}{DO}$
By applying the property of cross-multiplication (alternendo), we can rewrite this as:
$\frac{AO}{CO} = \frac{BO}{DO}$ --- (Equation 1)
Step 2: Construction
Draw a line $EO$ parallel to $AB$ such that $E$ lies on $AD$.
Construction: Draw $EO \parallel AB$, where $E$ is a point on $AD$.
Step 3: Applying Thales' Theorem (Basic Proportionality Theorem)
In $\triangle DAB$, since $EO \parallel AB$ [By construction], by the Basic Proportionality Theorem (BPT), which states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio:
$\frac{DE}{EA} = \frac{DO}{OB}$ --- (Equation 2)
Step 4: Comparing Equations
From Equation 1, we have:
$\frac{BO}{DO} = \frac{AO}{CO}$
Taking the reciprocal of both sides:
$\frac{DO}{BO} = \frac{CO}{AO}$ --- (Equation 3)
Comparing Equation 2 and Equation 3:
Since $\frac{DE}{EA} = \frac{DO}{OB}$ and $\frac{DO}{OB} = \frac{CO}{AO}$, it follows that:
$\frac{DE}{EA} = \frac{CO}{AO}$
Step 5: Conclusion using Converse of BPT
In $\triangle ADC$, we have $\frac{DE}{EA} = \frac{DO}{OC}$ (rearranging the terms from the previous step).
By the Converse of the Basic Proportionality Theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Therefore, $EO \parallel DC$.
Since we constructed $EO \parallel AB$ and we proved $EO \parallel DC$, it implies that $AB \parallel DC$.
Since one pair of opposite sides is parallel, the quadrilateral $ABCD$ is a trapezium.
Final Answer: Since $AB \parallel DC$, the quadrilateral $ABCD$ is a trapezium.
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 $\frac{AM}{AB} = \frac{AN}{AD}$.
- 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).
- Q9: $ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
CBSE Solutions for Class 10 Mathematics Triangles
Chapters in CBSE - Class 10 Mathematics
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