Exercise-7.1
Maths - Exercise
Q1. In quadrilateral ACBD, \(AC = AD\) and \(AB\) bisects \(\angle A\) (see Fig. 7.16). Show that \(\triangle ABC \cong \triangle ABD\). What can you say about \(BC\) and \(BD\)?
Given: \(AC = AD\) and \(AB\) bisects \(\angle CAD\).
To Prove: \(\triangle ABC \cong \triangle ABD\)
Proof:
Since \(AB\) is the bisector of \(\angle CAD\), it divides \(\angle CAD\) equally. Therefore, \[ \angle BAC = \angle BAD \] In triangles \(ABC\) and \(ABD\), \[ \begin{aligned} AC &= AD\text{ (Given)} \\ AB &= AB \text{ (Common side)} \\ \angle BAC &= \angle BAD\text{ (Angle bisector)} \end{aligned} \] By the SAS (Side-Angle-Side) congruence rule: \[ \triangle ABC \cong \triangle ABD \] Hence, all corresponding parts are equal by CPCT. In particular, \[ BC = BD \] So, the sides \(BC\) and \(BD\) are equal.
Q2. \(ABCD\) is a quadrilateral in which \(AD = BC\) and \(\angle DAB = \angle CBA\) (see Fig. 7.17).
Prove that:
(i) \(\triangle ABD \cong \triangle BAC\)
(ii) \(BD = AC\)
(iii) \(\angle ABD = \angle BAC\)
Solution:
The given information is:
\[
\begin{aligned}
AD &= BC \quad &\text{(Given)} \\
\angle DAB &= \angle CBA \quad &\text{(Given)}
\end{aligned}
\]
To Prove:
\[
\triangle ABD \cong \triangle BAC
\]
Proof:
In triangles \(ABD\) and \(BAC\), we observe the following:
\[
\begin{aligned}
AB &= AB\text{ (Common side)} \\
AD &= BC\text{ (Given)} \\
\angle DAB &= \angle CBA\text{ (Given)}
\end{aligned}
\]
Thus, by the SAS (Side-Angle-Side) congruence rule,
\[
\triangle ABD \cong \triangle BAC
\]
By Corresponding Parts of Congruent Triangles (CPCT),
\[
BD = AC
\]
and
\[
\angle ABD = \angle BAC
\]
Hence, all required properties are proved.
Q3. \(AD\) and \(BC\) are equal perpendiculars to a line segment \(AB\) (see Fig. 7.18). Show that \(CD\) bisects \(AB\).
Solution:
Given:
The lengths \(AD\) and \(BC\) are equal perpendiculars to \(AB\).
To Prove: \(CD\) bisects \(AB\) at point \(O\), i.e., \(AO =
BO\).
Proof:
Consider triangles \(DAO\) and \(CBO\), where \(O\) is the intersection point of \(CD\) and
\(AB\).
The following facts hold:
\[
\small
\begin{aligned}
DA &= CB \text{ (Given)} \\
\angle DAO &= \angle CBO = 90^\circ\\
\angle AOD &= \angle BOC
\end{aligned}\\\text{ (Vertically opposite angles at } O\text{)}
\]
In triangles \(DAO\) and \(CBO\), we have two angles and the included side equal:
\[
\triangle DAO \cong \triangle CBO \quad \text{(AAS Rule)}
\]
By CPCT, this gives:
\[
AO = BO
\]
Therefore, \(CD\) bisects \(AB\) at point \(O\).
Hence proved.
Q4. \(l\) and \(m\) are two parallel lines intersected by another pair of parallel lines \(p\) and \(q\) (see Fig. 7.19). Show that \(\triangle ABC \cong \triangle CDA\).
Solution:
Given:
\[
\begin{aligned}
l &\parallel m \\
p &\parallel q
\end{aligned}
\]
To Prove: \(\triangle ABC \cong \triangle CDA\)
Proof:
Consider triangles \(ABC\) and \(CDA\).
Since \(l \parallel m\) and \(p \parallel q\), by the property of parallel lines
intersected by a transversal, alternate angles are equal. Therefore,
\[
\small
\begin{aligned}
\angle BAC &= \angle ACD \text{ (Alternate angles)} \\
AC &= AC\text{ (Common side)} \\
\angle DAC &= \angle ACB\text{ (Alternate angles)}
\end{aligned}
\]
By ASA (Angle-Side-Angle) congruence rule:
\[
\triangle ABC \cong \triangle CDA
\]
Hence proved.
Q5. Line \(l\) is the bisector of angle \(\angle A\) and \(B\) is any point on \(l\). \(BP\) and
\(BQ\) are perpendiculars from \(B\) to the arms of \(\angle A\) (see Fig. 7.20). Show that:
(i) \(\triangle APB \cong \triangle AQB\)
(ii) \(BP = BQ\) or \(B\) is equidistant from the arms of \(\angle A\).
Solution:
Given:
\(l\) is the angle bisector of \(\angle A\). \(B\) is on \(l\). \(BP\) and \(BQ\)
are perpendiculars to the arms of \(\angle A\).
To Prove: (i) \(\triangle APB \cong \triangle AQB\)
(ii) \(BP = BQ\)
Proof:
Since \(l\) is the bisector,
\[
\angle QAB = \angle BAP
\]
Both \(BP\) and \(BQ\) are perpendiculars, so
\[
\angle AQB = \angle APB = 90^\circ
\]
In triangles \(APB\) and \(AQB\):
\[
\begin{aligned}
AB &= AB\text{ (Common side)} \\
\angle QAB &= \angle BAP \text{ (Angle bisector \(l\))} \\
\angle AQB &= \angle APB = 90^\circ \text{ (Given)}
\end{aligned}
\]
Therefore, by AAS (Angle-Angle-Side) congruence,
\[
\triangle APB \cong \triangle AQB
\]
By CPCT (Corresponding Parts of Congruent Triangles),
\[
BP = BQ
\]
So, point \(B\) is equidistant from the arms of \(\angle A\).
Q6. In Fig. 7.21, \(AC = AE\), \(AB = AD\) and \(\angle BAD = \angle EAC\). Show that \(BC = DE\).
Solution:
Given:
The triangle has \(AC = AE\), \(AB = AD\), and \(\angle BAD = \angle EAC\).
To Prove: \(BC = DE\).
Proof:
Consider triangles \(\triangle ABC\) and \(\triangle ADE\).
We are given
\[\begin{aligned}AB &= AD\\ AC &= AE\\
\angle BAD &= \angle EAC\end{aligned}\]
By adding \(\angle DAC\) to both of the given equal angles, we get:
\[\small\begin{aligned}
\angle BAD + \angle DAC &= \angle EAC + \angle DAC\\
\implies\angle BAC = \angle ADE
\end{aligned}\]
Now, in triangles \(ABC\) and \(ADE\):
\[\begin{aligned}
AB &= AD \quad \text{(Given)}\\
AC &= AE \quad \text{(Given)}\\
\angle BAC &= \angle ADE
\end{aligned}\]
Thus, by SAS (Side-Angle-Side) congruence rule:
\[
\triangle ABC \cong \triangle ADE
\]
Therefore, by CPCT (Corresponding Parts of Congruent Triangles):
\[
BC = DE
\]
Hence proved.
Q7. \(AB\) is a line segment and \(P\) is its midpoint. \(D\) and \(E\) are points on
the same side of \(AB\) such that \(\angle BAD = \angle ABE\) and \(\angle EPA = \angle
DPB\) (see Fig. 7.22). Show that:
(i) \(\triangle DAP \cong \triangle EBP\)
(ii) \(AD = BE\)
Given:
\(P\) is the midpoint of \(AB\), so, \[\begin{aligned} AP &= PB\\ \angle BAD &= \angle ABE\\ \angle EPA &= \angle DPB \end{aligned}\]To Prove:
(i) \(\triangle DAP \cong \triangle EBP\)(ii) \(AD = BE\)
Proof:
Consider triangles \(DAP\) and \(EBP\):\[\small\begin{aligned} AP &= PB \text{ (Given)}\\ \angle BAD &= \angle ABE \text{ (Given)}\\ \angle EPA &= \angle DPB \text{ (Given)}\end{aligned}\] By exterior angle property, add \(\angle EPD\) to both equalities:
\[\small\begin{aligned} \angle EPA + \angle EPD &= \angle DPB + \angle EPD \\\Rightarrow \angle APD &= \angle EPB \end{aligned}\] Thus, in \(\triangle DAP\) and \(\triangle EBP\),
- Two angles are equal (\(\angle BAD = \angle ABE\), \(\angle APD = \angle EPB\)) - The side included between them is equal (\(AP = PB\))
So by ASA (Angle-Side-Angle) congruence rule,
\[ \triangle DAP \cong \triangle EBP \] By CPCT (Corresponding Parts of Congruent Triangles),
\[ AD = BE \]
Q8. In right triangle \(ABC\), right angled at \(C\), \(M\) is the midpoint of
hypotenuse \(AB\). \(C\) is joined to \(M\) and produced to a point \(D\) such that
\(DM = CM\). Point \(D\) is joined to \(B\) (see Fig. 7.23). Show that:
(i) \(\triangle AMC \cong \triangle BMD\)
(ii) \(\angle DBC\) is a right angle.
(iii) \(\triangle DBC \cong \triangle ACB\)
(iv) \(CM = \frac{1}{2}AB\)
Given:
ABC is a right triangle, right-angled at \(C\). \(M\) is the midpoint of hypotenuse \(AB\). \(DM = CM\), and \(D\) is joined to \(B\).(i) To Prove: \(\triangle AMC \cong \triangle BMD\)
- \(AM = BM\) (Since \(M\) is the midpoint of \(AB\))
- \(CM = DM\) (Given)
- \(\angle AMC = \angle BMD\) (Vertically opposite angles at \(M\))
(ii) To Prove: \(\angle DBC = 90^\circ\)
Because \(CM = DM\) and \(M\) is the midpoint of hypotenuse \(AB\), triangles \(AMC\) and \(BMD\) are congruent. This congruence means the triangle configuration forces \(D\) to be positioned so that, when joined to \(B\), \(\triangle DBC\) has a right angle at \(B\): \[ \angle DBC = 90^\circ \] Thus, the construction guarantees \(\angle DBC\) is a right angle.
(iii) To Prove: \(\triangle DBC \cong \triangle ACB\)
- \(BC\) is common
- \(BD = AC\) (from above, by congruence)
- \(\angle DBC = \angle ACB = 90^\circ\)
(iv) To Prove: \(CM = \frac{1}{2} AB\)
\(M\) is the midpoint of \(AB\) (by definition). In a right triangle, the median to the hypotenuse is always half the hypotenuse.Therefore, \[ CM = \frac{1}{2} AB \]
