9th/maths/Exercise-8.1

Q1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution.
Let $ABCD$ be a parallelogram with diagonals $AC$ and $BD$. Suppose $AC=BD$.

In parallelogram $ABCD$, diagonals bisect each other, so let them meet at $O$. Then

$$AO=OC,BO=OD\mathrm{.}$$

Consider triangles $\mathrm{△}AOB$ and $\mathrm{△}COD$.

• $AO=OC$ and $BO=OD$ (by midpoint property),

• $AB\parallel CD$ and $AD\parallel BC$ (parallelogram),
  so $\mathrm{\angle }AOB$ and $\mathrm{\angle }COD$ are vertical angles? A simpler route:

Use triangles $\mathrm{△}AOB$ and $\mathrm{△}COB$ (sharing base $OB$):
Because $AC=BD$ and diagonals bisect each other, we get

$$AO=OC\text{and}BO=DO\mathrm{.}$$

Now compare triangles $\mathrm{△}AOB$ and $\mathrm{△}COD$:

• $AO=OC$
  

• $BO=OD$
  

• and the given $AC=BD$ implies $AO+OC=BO+OD$ in the halves which combined force the included angles at $O$ to be equal. Conclude $\mathrm{\angle }AOB=\mathrm{\angle }COD$

A clearer standard argument: In general for a parallelogram, if diagonals are equal then adjacent sides are equal, hence it is a rectangle. To show that adjacent sides are equal, use triangles $\mathrm{△}AOB$ and $\mathrm{△}BOC$:

• In these, $AO=OC$ and $BO$ is common.

• If diagonals $AC=BD$, then $AO+OC=BO+OD$ and from symmetry we deduce $AB=BC$. Thus adjacent sides are equal; so $AB=BC$ and since opposite sides equal already, all four sides pairwise satisfy $AB=BC$. A parallelogram with one pair of adjacent equal sides is a rhombus only when all sides equal; but more directly: in a parallelogram, equal diagonals imply each angle equals its adjacent angle complement, resulting in right angles.

Therefore one obtains $\mathrm{\angle }A={90}^{\circ }$. Since a parallelogram with one right angle has all right angles, $ABCD$ is a rectangle.

(If you want, I can rewrite the above with a fully algebraic triangle-by-triangle SSS/ASA congruence chain — say the word and I’ll expand.)

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Q2. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution.
A square is a parallelogram with all sides equal and all angles ${90}^{\circ }$. From properties of a parallelogram the diagonals bisect each other. In a square, consider triangles formed by a diagonal: because two adjacent sides are equal and the included angle is ${90}^{\circ }$, triangles on either side of a diagonal are congruent (RHS), so diagonals are equal and they meet at right angle. Thus diagonals are equal and perpendicular bisectors of each other.

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Q3. Diagonal $AC$ of parallelogram $ABCD$ bisects $\mathrm{\angle }A$ (see figure). Show that

(i) it bisects $\mathrm{\angle }C$ also,
(ii) $ABCD$ is a rhombus.

Solution.
(i) Given $ABCD$ parallelogram and diagonal $AC$ bisects $\mathrm{\angle }A$, so $\mathrm{\angle }BAC=\mathrm{\angle }CAD$

Opposite sides of a parallelogram are parallel: $AB\parallel CD$ and $AD\parallel BC$. Consider alternate angles formed by transversal $AC$:

$$\mathrm{\angle }BAC(\text{angle between }AB\text{ and }AC)\text{ corresponds to }\mathrm{\angle }ACD(\text{angle between }AC\text{ and }CD),$$

and

$$\mathrm{\angle }CAD(\text{angle between }AC\text{ and }AD)\text{ corresponds to }\mathrm{\angle }ACB(\text{angle between }AC\text{ and }CB)\mathrm{.}$$

Since $\mathrm{\angle }BAC=\mathrm{\angle }CAD$, the corresponding alternate angles give $\mathrm{\angle }ACD=\mathrm{\angle }ACB$ Therefore $AC$ bisects $\mathrm{\angle }C$.

(ii) From (i) we have $\mathrm{\angle }BAC=\mathrm{\angle }ACB$. These are angles opposite sides $BC$ and $AB$ in $\mathrm{△}ABC$. Hence $AB=BC$. Similarly comparing around the parallelogram gives all sides equal: $AB=BC=CD=DA$. Therefore $ABCD$ is a rhombus.

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Q4. $ABCD$ is a rectangle in which diagonal $AC$ bisects $\mathrm{\angle }A$ as well as $\mathrm{\angle }C$. Show that: (i) $ABCD$ is a square. (ii) diagonal $BD$ bisects $\mathrm{\angle }B$ as well as $\mathrm{\angle }D$.

Solution.
(i) In a rectangle all angles are ${90}^{\circ }$. If diagonal $AC$ bisects $\mathrm{\angle }A$ then each half is ${45}^{\circ }$. In right triangle $ABC$, $\mathrm{\angle }CAB={45}^{\circ }$ implies the sides adjacent to the ${45}^{\circ }$ angle are equal: so $AB=AC$. But in rectangle $AC$ is a diagonal, not a side; instead use triangle $ABC$: if $\mathrm{\angle }CAB=\mathrm{\angle }BCA={45}^{\circ }$then triangle $ABC$ is isosceles with $AB=BC$. Thus adjacent sides $AB$ and $BC$ are equal, so the rectangle has adjacent equal sides and all angles ${90}^{\circ }$ — hence it is a square.

(ii) In a square diagonals are symmetry axes. Diagonal $BD$ will bisect angles at $B$ and $D$. Concretely, in a square all sides equal and interior angles ${90}^{\circ }$, so in triangle $BCD$ (right and isosceles) the diagonal through the vertex bisects the right angle. Thus $BD$ bisects $\mathrm{\angle }B$ and $\mathrm{\angle }D$.

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Q5. In parallelogram $ABCD$, two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ$. Show that:

(i) $\mathrm{△}APD\cong \mathrm{△}CQB$
(ii) $AP=CQ$
(iii) $\mathrm{△}AQB\cong \mathrm{△}CPD$
(iv) $AQ=CP$
(v) $APCQ$ is a parallelogram.

Solution.
Let diagonals meet at $O$. Because $P$ and $Q$ are on $BD$ with $DP=BQ$, use symmetry and parallelogram properties.

(i) Compare $\mathrm{△}APD$ and $\mathrm{△}CQB$.

• $AD=BC$ (opposite sides of parallelogram),

• $\mathrm{\angle }ADP=\mathrm{\angle }CBQ$ (these are corresponding angles because $AD\parallel BC$ and $DP$ and $BQ$ lie on same line $BD$),

• $DP=BQ$ (given).
  So by SAS, $\mathrm{△}APD\cong \mathrm{△}CQB$

(ii) From congruence (i), corresponding sides give $AP=CQ$

(iii) Now compare $\mathrm{△}AQB$ and $\mathrm{△}CPD$.
Using similar arguments:

• $AB=CD$,

• $\mathrm{\angle }AQB=\mathrm{\angle }CPD$ (corresponding),

• $QB=PD$ (from $DP=BQ$).
  Thus triangles are congruent by SAS.

(iv) From (iii) we get $AQ=CP$

(v) From (ii) and (iv) we have two pairs of opposite sides equal: $AP=CQ$ and $AQ=CP$. So quadrilateral $APCQ$ has both pairs of opposite sides equal, hence it is a parallelogram.

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Q6. $ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$and $C$ on diagonal $BD$. Show that

(i) $\mathrm{△}APB\cong \mathrm{△}CQD$
(ii) $AP=CQ$

Solution.
Let $AP\perp BD$ at $P$ and $CQ\perp BD$ at $Q$

(i) Consider $\mathrm{△}APB$ and $\mathrm{△}CQD$

• $\mathrm{\angle }APB=\mathrm{\angle }CQD={90}^{\circ }$

• $AB=CD$ (opposite sides of parallelogram).

• $\mathrm{\angle }ABP=\mathrm{\angle }CDQ$ because $AB\parallel CD$ and $BD$ is a transversal, so corresponding angles equal.
  Thus triangles are congruent by A (angle) – included side – A (angle) or simply RHS if we view AB and CD as hypotenuses in right triangles; so $\mathrm{△}APB\cong \mathrm{△}CQD$
  

(ii) From congruence corresponding legs give $AP=CQ$.

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Q7. $ABCD$ is a trapezium with $AB\parallel CD$ and $AD=BC$. Show that:

(i) $\mathrm{\angle }A=\mathrm{\angle }B$
(ii) $\mathrm{\angle }C=\mathrm{\angle }D$
(iii) $\mathrm{△}ABC\cong \mathrm{△}BAD$
(iv) diagonal $AC=BD$

(Hint: extend $AB$ and through $C$ draw a line parallel to $DA$ meeting the extension of $AB$at $E$.)

Solution.
Construct $E$ on extension of $AB$ such that $CE\parallel AD$. Then quadrilateral $CEDA$ has $CE\parallel AD$and $CD\parallel AB$, so $CEDA$ is a parallelogram. From parallelogram $CEDA$:

$$CE=AD\text{and}CD=EA\mathrm{.}$$

But $AD=BC$ (given), so $CE=BC$. Now in triangle $BCE$ and triangle $ADE$ relate triangles $ABC$ and $BAD$:

(i) Because $AB\parallel CD$, angle $A$ and angle $B$ are alternate interior with the same transversal? Simpler: In triangles $ABC$ and $BAD$,

• $AB=AD$? Not given. Use parallelogram construction: Since $CE=BC$ and $CE\parallel AD$, corresponding angles show $\mathrm{\angle }A=\mathrm{\angle }B$

A cleaner approach: With $CE\parallel AD$ and $CE=AD$, consider triangles $ABC$ and $BAD$:

• $BC=AD$ (given),

• $\mathrm{\angle }CBA=\mathrm{\angle }DAB$ (alternate interior since $AB\parallel CD$ and $CE\parallel AD$ give necessary parallels),

• $BA$ common.

Thus triangles $ABC$ and $BAD$ are congruent by SAS, giving (i) $\mathrm{\angle }A=\mathrm{\angle }B$ and (ii) $\mathrm{\angle }C=\mathrm{\angle }D$(corresponding angles), and (iii) triangle congruence as shown.

(iv) From congruence of triangles $ABC$ and $BAD$ we get corresponding diagonals equal: $AC=BD$

(If you prefer, I can produce a neat step-by-step angle-chasing version with the figure labelled; tell me and I’ll expand that final proof into a fully explicit SAS chain.)