Q1. ABCD is a quadrilateral in which P, Q, R and S are
mid-points of the sides AB, BC, CD and DA
(see Fig 8.20). AC is a diagonal. Show that :
(i) \(SR || AC\) and \(SR = \frac{1}{2}AC\)
(ii) \(PQ = SR\)
(iii) \(PQRS\) is a parallelogram.
Solution:
Given:\(ABCD \text{ and } \;P,\;Q,\;R,\;S\) are the mid-points of\(AB,\;BC,\;CD,\;DA\) respectively. Diagonal \( AC \)is drawn.To prove:
\[SR∥AC \text{ and, }\\ SR=\frac{1}{2}AC\\ PQ=SR\\ PQRS\text{ is a parallelogram}\].
(i)Proof
\[ SR \parallel AC \]
\[ SR = \frac{1}{2}AC \]
In triangle \(ACD\), points \(S\) and \(R\) are the mid-points of sides \(AD\) and \(CD\). By the Midpoint Theorem, the segment joining the mid-points of two sides of a triangle is parallel to the third side and half of its length. Therefore, \[ SR \parallel AC \] and \[ SR = \frac{1}{2}AC. \]
(ii)Proof
\[ PQ = SR \]
In triangle \(ABC\), points \(P\) and \(Q\) are the mid-points of \(AB\) and \(BC\). By the Midpoint Theorem: \[ PQ = \frac{1}{2}AC. \] From part (i), \[ SR = \frac{1}{2}AC. \] Thus, \[ PQ = SR. \]
(iii)Proof
\[ PQRS \text{ is a parallelogram} \]
Since \(PQ = SR\) and \(PQ \parallel SR\), one pair of opposite sides is equal and parallel. Hence, \(PQRS\) is a parallelogram.
Additionally, in triangle \(DAB\), midpoints \(S\) and \(P\) give: \[ SP = \frac{1}{2}DB, \quad SP \parallel DB. \] In triangle \(DCB\), midpoints \(R\) and \(Q\) give: \[ RQ = \frac{1}{2}DB, \quad RQ \parallel DB. \] Thus, \[ SP = RQ \quad\text{and}\quad SP \parallel RQ. \] Therefore, both pairs of opposite sides are equal and parallel, confirming that \[ PQRS \text{ is a parallelogram}. \]
Q2. \(ABCD\) is a rhombus and \(P,\; Q,\; R\text{ and }S\) are the mid-points of the sides \(AB,\; BC,\; CD\text{ and }DA\) respectively. Show that the quadrilateral \(PQRS\) is a rectangle.
Solution:
Given:
\(ABCD\) is a rhombus and \(P,\; Q,\; R\text{ and }S\) are the mid-points of the sides \(AB,\; BC,\; CD\text{ and }DA\) respectively
(i)Proof
In Δ ABD, P and S are the mid-points of AB and AD. By the Midpoint Theorem: \[ PS = \frac{1}{2}BD \] \[ PS \parallel BD \tag{1}\]
In Δ CBD, Q and R are the mid-points of BC and CD. By the Midpoint Theorem: \[ QR = \frac{1}{2}BD \] \[ QR \parallel BD \tag{2}\]
From (1) and (2): \[ PS = QR \] \[ PS \parallel QR \]
Similarly, in Δ ABC: \[ PQ = \frac{1}{2}AC \] \[ PQ \parallel AC \] and in Δ ACD: \[ SR = \frac{1}{2}AC \] \[ SR \parallel AC \]
Thus: \[ PQ = SR \] Opposite sides of quadrilateral PQRS are equal and parallel. Therefore PQRS is a parallelogram.
(ii) Using Rhombus Property
Diagonal of a rhombus are perpendicular, so: \[ AC \perp BD \]
Since \[ PQ \parallel AC \] and \[ PS \parallel BD \] we get: \[ PQ \perp PS \]
Similarly, \[ SR \parallel AC \] and \[ QR \parallel BD \] give: \[ SR \perp QR \]
Thus all angles of PQRS are right angles: \[ \angle P = \angle Q = \angle R = \angle S = 90^\circ \]
Hence, \[ PQRS \text{ is a rectangle.} \]
Q3. \(ABCD\) is a rectangle and \(P,\; Q,\; R\text{ and }S\) are mid-points of the sides \(AB,\; BC,\; CD\text{ and }DA\) respectively. Show that the quadrilateral \(PQRS\) is a rhombus.
Solution:
\(ABCD\) is a rectangle and \(P,\; Q,\; R\text{ and }S\) are mid-points of the sides \(AB,\; BC,\; CD\text{ and }DA\) respectively.
(i) Showing that opposite sides of PQRS are equal and parallel
Consider Δ ADC. Since S and R are the mid-points of AD and CD, by the Midpoint Theorem: \[ SR = \frac{1}{2}AC \] \[ SR \parallel AC \tag{1}\]
Consider Δ ABC. P and Q are the mid-points of AB and BC, therefore by the Midpoint Theorem: \[ PQ = \frac{1}{2}AC \] \[ PQ \parallel AC \tag{2}\]
From (1) and (2): \[ SR = PQ \] \[ SR \parallel PQ \] So one pair of opposite sides of PQRS are equal and parallel.
Similarly, considering Δ ABD and Δ CDB: P and S are mid-points of AB and AD; Q and R are mid-points of BC and CD. By the Midpoint Theorem: \[ SP = \frac{1}{2}BD \] \[ SP \parallel BD \] and \[ QR = \frac{1}{2}BD \] \[ QR \parallel BD \] Thus: \[ SP = QR \] \[ SP \parallel QR \]
Hence both pairs of opposite sides of PQRS are equal and parallel. Therefore: \[ PQRS \text{ is a parallelogram}. \]
(ii) Using properties of a rectangle to show PQRS is a rhombus
In a rectangle: \[ AC = BD \quad \text{(diagonals are equal)} \]
But in PQRS we already proved: \[ PQ = SR = \frac{1}{2}AC \] \[ SP = QR = \frac{1}{2}BD \] Since: \[ AC = BD \] we get: \[ PQ = QR = RS = SP \] So all four sides of PQRS are equal.
A parallelogram with all sides equal is a rhombus. Therefore: \[ PQRS \text{ is a rhombus}. \]
Q4. ABCD is a trapezium in which AB || DC, BD is a diagonal and \(E\) is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that F is the mid-point of BC.
Solution:
ABCD is a trapezium in which \(AB ∥ DC.\; BD\) is a diagonal and \(E\) is the mid-point of \(AD.\) A line through \(E\) parallel to \(AB\) meets \(BC\) at \(F.\)
(i) Using midpoint theorem in \(\triangle ABD\)
In \(\triangle ABD\), \(E\) is the mid-point of \(AD\). A line through \(E\) is drawn parallel to \(AB\), therefore: \[ EF \parallel AB \] By the Midpoint Theorem, if a line is drawn through the mid-point of one side of a triangle parallel to another side, it bisects the third side. So F is the mid-point of: \[ DF \text{ on } BD \]
(ii) Using midpoint theorem in \(\triangle DBC\)
Now consider \(\triangle DBC\). We already know that \(D\) is a vertex and \(F\) lies on \(BC\) such that: \[ EF \parallel DC \quad \text{(because } AB \parallel DC \text{ and } EF \parallel AB) \] Since E is the mid-point of AD, and ED is a part of triangle DBC, using midpoint theorem in \(\triangle DBC\): \[ F \text{ is the mid-point of } BC \]
Therefore:
\[ F \text{ is the mid-point of } BC. \]
Q5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.22). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
\(ABCD\) is a parallelogram. \(E\) and \(F\) are the mid-points of \(AB\) and \(CD\) respectively.
(i) Showing \(AECF\) is a parallelogram
In quadrilateral \(AECF,\) \(E\) is the mid-point of \(AB\) and \(F\) is the mid-point of \(CD.\) Since \(\)AB ∥ CD (opposite sides of parallelogram) we get: \[ AE \parallel FC \] Also \(AD ∥ BC\), so: \[ AF \parallel EC \] Therefore AECF is a parallelogram.
(ii) Using \(\triangle DQC\) to find midpoint \(P\) on \(BD\)
Let AF intersect BD at P. In \(\triangle DCQ,\; F\) is the mid-point of \(CD\). Since (AF ∥ EC\) (from above), \(AF\) acts as a mid-line in \(\triangle DCB.\) Thus by the Midpoint Theorem: \[ DP = PQ \tag{1}\]
(iii) Using \(\triangle ABP\) to find midpoint \(Q\) on \(BD\)
Let EC intersect BD at Q. \(E\) is mid-point of \(AB\) and \(EC ∥ AF.\) In \(\triangle ABP,\; E\) is mid-point of \(AB\) and \(EC ∥ AP.\) So by the Midpoint Theorem: \[ PQ = QB \tag{2}\]
(iv) Combining results
From (1) and (2): \[ DP = PQ = QB \] Therefore BD is divided into three equal parts: \[ DP = PQ = QB \] So AF and EC trisect the diagonal BD.
Hence:
\[ AF \text{ and } EC \text{ trisect diagonal } BD. \]
Q6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB
and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) \(CM = MA = \frac{1}{2} AB\)
Solution:
ABC is a triangle right angled at C. M is the mid-point of hypotenuse AB. A line through M parallel to BC meets AC at D.
(i)Proving D is the mid-point of AC
Since the line through M is drawn parallel to BC, we have: \[ MD \parallel BC. \] Note that \(A,\;M,\;B\) are collinear and \(A,\;D,\;C\) are collinear. Compare triangles \( \triangle AMD\) and \( \triangle ABC\). Because \(AM\) lies on \(AB\) and \(AD\) lies on \(AC\), and \(MD\parallel BC\), the corresponding angles are equal: \[ \angle AMD = \angle ABC,\qquad \angle ADM = \angle ACB. \] Hence \( \triangle AMD \sim \triangle ABC\). From this similarity we get \[ \frac{AM}{AB}=\frac{AD}{AC}. \] But \(AM\) is half of \(AB\) (since \(M\) is the mid-point of \(AB\)), so \(\dfrac{AM}{AB}=\dfrac{1}{2}\). Therefore \[ \frac{AD}{AC}=\frac{1}{2}\quad\Rightarrow\quad AD=\frac{1}{2}AC. \] Thus \(D\) is the mid-point of \(AC\).
(ii) Proving \(MD\perp AC\)
Since \(ABC\) is right angled at \(C\), we have \[ BC\perp AC. \] But \(MD\parallel BC\). Therefore \[ MD\perp AC. \] So \(MD\) is perpendicular to \(AC\).
(iii)Proof: \[CM=MA=\dfrac{1}{2}AB\]
Because \(\angle ACB=90^\circ\), point \(C\) lies on the circle with diameter \(AB\) (angle in a semicircle is a right angle). The centre of this circle is the midpoint \(M\) of the diameter \(AB\). Hence \(M\) is equidistant from \(A,\;B\) and \(C\). Therefore \[ CM = MA = MB. \] But \(MA=MB=\dfrac{1}{2}AB\) (since \(M\) is the mid-point of \(AB\)). Thus \[ CM = MA = \frac{1}{2}AB. \]
Conclusion:
(i) \(D\) is the mid-point of \(AC\).
(ii) \(MD\perp AC\).
(iii) \(CM = MA = \dfrac{1}{2}AB\).
