Detailed Solutions with Labelled Diagrams (Exercise Sets 12.1, 12.2 & 12.3)
Two sides of a quadrilateral are called adjacent sides if they have a common endpoint (vertex).
Side BC shares vertex B with side AB.
Side AD (or DA) shares vertex A with side AB.
∴ The sides adjacent to side AB are BC and AD.
Side CD (or DC) has no common vertex with side AB.
∴ The side opposite to side AB is CD.
Internal angles at adjacent vertices (endpoints of the same side) are called adjacent angles.
Vertex B lies on side AB with vertex A, and vertex D lies on side AD with vertex A.
∴ The angles adjacent to ∠A are ∠B and ∠D.
Vertex C is not adjacent to vertex A.
∴ The angle opposite to ∠A is ∠C.
Pair of opposite sides: Two sides of a quadrilateral that do not have a common endpoint (vertex); that is, two sides that are not adjacent to each other.
Pair of opposite angles: Two internal angles of a quadrilateral whose vertices are not the endpoints of the same side; that is, two internal angles that are not adjacent.
Let ABCD be a planar, non-self-intersecting quadrilateral.
Consider any vertex A, its two adjacent vertices B and D, and the opposite vertex C.
The two rays AB and AD form two angles at vertex A: one angle smaller than 180° and one reflex angle greater than 180°.
Case 1 (When vertex A does not form a dent):
If vertex A lies outside △BCD (so ray CA starting from C and passing through A does not intersect segment BD), then the internal angle at vertex A is the angle ∠BAD < 180°.
Case 2 (When vertex A forms a dent):
If vertex A lies inside △BCD (so ray CA starting from C and extended through A intersects segment BD), then the internal angle at vertex A is the reflex angle ∠BAD > 180° (equal to 360° − ∠BAC − ∠CAD).
Since ABCD is a non-self-intersecting quadrilateral and AB ∥ DC, sides AD and BC act as transversals between parallel lines AB and DC.
Interior angles on the same side of a transversal are supplementary:
∠A + ∠D = 180°
∠B + ∠C = 180°
Since each angle of a quadrilateral is greater than 0°, it follows that:
∠A < 180°, ∠B < 180°, ∠C < 180°, and ∠D < 180°.
∴ No, ABCD cannot be non-convex when AB ∥ DC; it is always convex.
Consider a non-convex dart ABCD (Fig 12.1.3(b)) where vertex D lies inside △ABC such that CD = AB.
Here AB = CD, but the internal angle at vertex D is a reflex angle (∠D > 180°).
∴ Yes, ABCD can be non-convex when AB = CD.
Consider a non-convex dart ABCD (Fig 12.1.3(b)) in which AB = CB and AD = CD, with D inside △ABC.
Join BD. In △ABD and △CBD:
AB = CB (By construction)
AD = CD (By construction)
BD = BD (Common side)
∴ △ABD ≅ △CBD (By SSS congruence rule)
∴ ∠A = ∠C (By CPCT)
Since vertex D lies inside △ABC, internal angle ∠D > 180°.
∴ Yes, ABCD can be non-convex when ∠A = ∠C.
In quadrilateral ABCD, the four sides are AB, BC, CD, and DA, and the diagonals are AC and BD.
1. When D lies in Region 2 (across side AC, opposite to vertex B):
Diagonals AC and BD intersect each other inside the quadrilateral.
∴ Quadrilateral ABCD is Convex.
2. When D lies in Region 1 (interior of △ABC) or Region 3 (vertically opposite to ∠B):
In Region 1, vertex D lies inside △ABC, making internal angle ∠D > 180° (dent at D).
In Region 3, vertex B lies inside △ADC, making internal angle ∠B > 180° (dent at B).
∴ In Regions 1 and 3, quadrilateral ABCD is Non-convex.
3. When D lies in Region 4, Region 5, Region 6, or Region 7:
In these four regions, either side CD crosses side AB or side DA crosses side BC.
∴ In Regions 4, 5, 6, and 7, quadrilateral ABCD is Self-intersecting.
Suppose a quadrilateral ABCD is self-intersecting.
Then two of its opposite sides (say AB and CD) must intersect at some point E.
Two intersecting lines (line AB and line CD) always lie in one unique plane.
Since vertices A and B lie on line AB, and vertices C and D lie on line CD, all four vertices A, B, C, and D must lie in that same plane.
∴ No, a quadrilateral cannot be both self-intersecting and non-planar.
Let ABCD be a parallelogram in which ∠A = 90°.
∠C = ∠A = 90° (Opposite angles of a parallelogram are equal)
∠A + ∠B = 180° (Interior angles on the same side of transversal AB as AD ∥ BC)
90° + ∠B = 180° ⇒ ∠B = 90°
∠D = ∠B = 90° (Opposite angles of a parallelogram are equal)
∴ Since ∠A = ∠B = ∠C = ∠D = 90°, ABCD is a rectangle.
The diagonals of every rhombus are perpendicular to each other.
However, a rhombus need not have 90° interior angles (for example, a rhombus with angles 60° and 120° is not a square).
Let ABCD be a parallelogram in which diagonal AC = diagonal BD.
In △ABC and △DCB:
AB = DC (Opposite sides of a parallelogram are equal)
BC = CB (Common side)
AC = BD (Given)
∴ △ABC ≅ △DCB (By SSS congruence rule)
∴ ∠ABC = ∠DCB (By CPCT)
Since AB ∥ DC and BC is a transversal:
∠ABC + ∠DCB = 180° (Co-interior angles)
∠ABC + ∠ABC = 180°
2∠ABC = 180° ⇒ ∠ABC = 90°
Since ABCD is a parallelogram with one right angle, all its angles are 90°.
∴ ABCD is a rectangle.
Given: ABCD is a parallelogram and diagonal AC bisects ∠A, so ∠1 = ∠2.
To Prove: (i) AC bisects ∠C (i.e., ∠3 = ∠4), and (ii) ABCD is a rhombus.
Part (i):
Since ABCD is a parallelogram, AB ∥ DC and AD ∥ BC.
Since AB ∥ DC and AC is a transversal:
∠1 = ∠3 (Alternate interior angles) — (1)
Since AD ∥ BC and AC is a transversal:
∠2 = ∠4 (Alternate interior angles) — (2)
Also, ∠1 = ∠2 (Given, AC bisects ∠A) — (3)
From (1), (2), and (3), we get: ∠3 = ∠4
∴ Diagonal AC also bisects ∠C.
Part (ii):
From (1) and (3), we have ∠2 = ∠3 (or ∠1 = ∠4).
In △ABC: ∠1 = ∠4 ⇒ AB = BC.
Also, AB = CD and BC = DA (Opposite sides of parallelogram).
∴ AB = BC = CD = DA.
Given that diagonal AC bisects ∠A and ∠C, and diagonal BD bisects ∠B and ∠D.
In △ABC and △ADC, ASA congruence shows △ABC ≅ △ADC.
∴ AB = AD and BC = DC (By CPCT) — (1)
Similarly, △ABD ≅ △CBD shows AB = CB and AD = CD — (2)
From (1) and (2), AB = BC = CD = DA, so ABCD must be a rhombus.
Let diagonals AC and BD intersect at point E at right angles such that EA = EC and EB = ED.
By SAS congruence in △AEB, △CEB, △CED, and △AED, all four triangles are congruent.
∴ AB = BC = CD = DA, so ABCD must be a rhombus.
A quadrilateral with equal diagonals need not be a rectangle (e.g., an isosceles trapezium).
Extra condition required: Add that ABCD is a parallelogram (or that the diagonals bisect each other).
Given: Parallelogram ABCD with AB ≠ BC. Angle bisectors intersect to form PQRS.
To Prove: PQRS is a rectangle.
Since ABCD is a parallelogram, ∠DAB + ∠CDA = 180°.
Dividing by 2: ½∠DAB + ½∠CDA = 90° ⇒ ∠SAD + ∠SDA = 90°.
In △ASD, ∠ASD = 180° − 90° = 90°.
∴ ∠PSR = ∠ASD = 90° (Vertically opposite angles).
Similarly, ∠PQR = 90°, ∠SPQ = 90°, and ∠SRQ = 90°.
Since all four interior angles are 90°, PQRS is a rectangle.
By the Midpoint Theorem:
PQ ∥ BC and PQ = ½BC = BR = RC — (1)
QR ∥ AB and QR = ½AB = AP = PB — (2)
PR ∥ AC and PR = ½AC = AQ = QC — (3)
1. In △PQR and △QPA: PQ = QP, QR = PA, PR = QA ⇒ △PQR ≅ △QPA (SSS).
2. In △PQR and △RBP: PQ = RB, QR = BP, PR = RP ⇒ △PQR ≅ △RBP (SSS).
3. In △PQR and △QRC: PQ = RC, QR = RQ, PR = QC ⇒ △PQR ≅ △QRC (SSS).
Thus, the two other congruent triangles are △RBP and △QRC.
Yes, draw straight lines through P parallel to QR, through Q parallel to PR, and through R parallel to PQ. Their intersections form vertices A, B, and C of △ABC.
Given: M is the midpoint of AB and N is the midpoint of AC. MN intersects AD at O.
To Prove: AO = OD.
In △ABC, M and N are midpoints of AB and AC ⇒ MN ∥ BC (Midpoint Theorem).
Since O lies on MN and D lies on BC, MO ∥ BD.
Now, in △ABD, M is the midpoint of AB and MO ∥ BD.
∴ O is the midpoint of AD (Converse of Midpoint Theorem).
∴ AO = OD.
Given: Quadrilateral $ABCD$ with $AB \parallel DC$ and $AB \neq CD$. $G$ and $H$ are midpoints of diagonals $AC$ and $BD$ respectively.
To Prove: $GH \parallel AB$.
Construction: Let $M$ be the midpoint of side $AD$. Join $MH$ and $MG$.
In $\triangle ABD$, $M$ is the midpoint of $AD$ and $H$ is the midpoint of $BD$. By the Midpoint Theorem, $MH \parallel AB$ — (1)
In $\triangle ADC$, $M$ is the midpoint of $AD$ and $G$ is the midpoint of $AC$. By the Midpoint Theorem, $MG \parallel DC$. Since $DC \parallel AB$ (Given), it follows that $MG \parallel AB$ — (2)
From (1) and (2), both line segments $MH$ and $MG$ pass through point $M$ and are parallel to $AB$. Since there is only one line through $M$ parallel to $AB$, points $M, H,$ and $G$ must lie on the same straight line.
∴ $GH \parallel AB$.
In △ABC, P and Q are midpoints of AB and BC ⇒ PQ ∥ AC and PQ = ½AC — (1)
In △ADC, S and R are midpoints of AD and CD ⇒ SR ∥ AC and SR = ½AC — (2)
From (1) and (2), PQ ∥ SR and PQ = SR, so PQRS is a parallelogram.
Since PR and QS are its diagonals, PR and QS bisect each other.
In △BCD, QR = ½BD — (3). Given AC = BD ⇒ PQ = QR.
In △POQ and △ROQ, PO = RO, OQ = OQ, and PQ = RQ ⇒ △POQ ≅ △ROQ (SSS).
∴ ∠POQ = ∠ROQ = 90° ⇒ PR ⊥ QS.
Converse: Yes, if PR ⊥ QS, then △POQ ≅ △ROQ (SAS), giving PQ = RQ, which implies AC = BD.
Construction: Use compass to measure AP and AS. Draw arcs intersecting at A'. Extend A'P to B' with PB' = A'P, B'Q to C' with QC' = B'Q, and A'S to D' with SD' = A'S.
Justification: By SSS and SAS congruence rules, △A'PS ≅ △APS, △B'PQ ≅ △BPQ, △D'SR ≅ △DSR, and △C'QR ≅ △CQR, proving A'B'C'D' ≅ ABCD.
By Midpoint Theorem, PQ = ½AC and QR = ½BD.
Direct: If PQRS is a square, PQ = QR and PQ ⊥ QR ⇒ AC = BD and AC ⊥ BD.
Converse: If AC = BD and AC ⊥ BD, then PQ = QR (rhombus) and PQ ⊥ QR (90° angle), making PQRS a square.