Q1. \(\mathrm{\triangle ABC \text{ and } \triangle DBC}\) are two isosceles triangles on
the same base BC and vertices A and D are on the
same side of BC (see Fig. 7.39). If AD is extended
to intersect BC at P, show that
(i) \(\mathrm{ \triangle ABD \cong \triangle ACD}\)
(ii) \(\mathrm{\triangle ABP \cong \triangle ACP}\)
(iii) AP bisects \(\mathrm{\angle A}\) as well as \(\mathrm{\angle D.}\)
(iv) AP is the perpendicular bisector of BC.
To prove:
\(\triangle ABD \cong \triangle ACD\)Given:
\[ \scriptsize \begin{aligned} AB &= AC \text{ (Sides of isosceles } \triangle ABC \text{)} \\ BD &= CD \text{ (Sides of isosceles } \triangle DBC \text{)} \end{aligned} \]Proof:
In \( \triangle ABD \) and \( \triangle ACD \): \[ \begin{aligned} AB &= AC \text{ (Given)} \\ BD &= CD \text{ (Given)} \\ AD &= AD \text{ (Common side)} \end{aligned} \] By SSS Congruence Rule: \[ \triangle ABD \cong \triangle ACD \]To prove:
\(\triangle ABP \cong \triangle ACP\)Proof:
In \( \triangle ABP \) and \( \triangle ACP \): \[ \begin{aligned} AB &= AC \text{(Given)} \\ AP &= AP \text{(Common side)} \\ \angle BAP &= \angle CAP \end{aligned} \] \[\text{(CPCT from } \triangle ABD \cong \triangle ACD )\] By SAS Congruence Rule: \[ \triangle ABP \cong \triangle ACP \]To show:
\[ \angle APB = \angle APC \] (from CPCT) Also, since points \(P, B, C\) are collinear, \[ \angle APB + \angle APC = 180^{\circ} \] (linear pair) Because \(\angle APB = \angle APC\), we have \[ 2 \angle APB = 180^{\circ} \implies \angle APB = 90^{\circ} \] \[ \boxed{ \triangle ABD \cong \triangle ACD \\ \triangle ABP \cong \triangle ACP \\ \angle APB = \angle APC = 90^{\circ} } \]
Q2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects \(\mathrm{\angle A}\).
Solution:
Given:
\[ AB = AC \]To Prove:
\( AD \) bisects \( BC \)Proof:
Consider triangles \( ABD \) and \( ACD \): \[ \begin{aligned} AB &= AC \text{ (Given)} \\ AD &= AD \text{ (Common side)} \\ \angle ABD &= \angle ACD \end{aligned} \\\text{ (Equal angles in isosceles } \triangle ABC \text{)} \] By SAS Congruence Rule: \[ \triangle ABD \cong \triangle ACD \] By CPCT (Corresponding Parts of Congruent Triangles): \[ \begin{aligned} BD &= CD \\ \angle BAD &= \angle CAD \end{aligned} \] Therefore, \( AD \) bisects \( BC \) and \( AD \) bisects \( \angle A \). Hence proved.
Q3. Two sides AB and BC and median AM
of one triangle ABC are respectively
equal to sides PQ and QR and median
PN of ∆ PQR (see Fig. 7.40). Show that:
(i) \(\mathrm{\triangle ABM \cong \triangle PQN}\)
(ii) \(\mathrm{\triangle ABC \cong \triangle PQR}\)
Solution:
Given:
Triangles \( \triangle ABC \) and \( \triangle PQR \) \[ \begin{aligned} AB &= PQ \\ BM &= QN \\ AM &= PN \\ \end{aligned} \] where \( AM \) and \( PN \) are medians.To Prove:
\[ \triangle ABM \cong \triangle PQN \]Proof:
In \( \triangle ABM \) and \( \triangle PQN \): \[ \begin{aligned} AB &= PQ \quad \text{(Given)} \\ BM &= QN \quad \text{(Given)} \\ AM &= PN \quad \text{(Given)} \end{aligned} \] By SSS (Side-Side-Side) Congruence Rule: \[ \triangle ABM \cong \triangle PQN \]To Prove:
\[ \triangle ABC \cong \triangle PQR \]Reasoning:
Since \( AM \) and \( PN \) are medians, \[ BC = 2 \times BM \\ QR = 2 \times QN \] By CPCT from \( \triangle ABM \cong \triangle PQN \): \[ BM = QN \implies BC = QR \] Also, \( AB = PQ \) (Given), Angles at \( B \) and \( Q \) are equal (from CPCT). Thus, by SAS (Side-Angle-Side) Congruence Rule: \[ \triangle ABC \cong \triangle PQR \] Hence proved.
Q4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
To Prove:
\( \triangle ABC \) is an isosceles triangle.Proof:
Consider triangles \( \triangle BFC \) and \( \triangle CEB \): \[ \begin{aligned} BC &= BC \text{ (Hypotenuse, common side)} \\ CF &= BE \text{ (Given)} \\ \angle F &= \angle E = 90^\circ \text{ (Right angles)} \end{aligned} \] By RHS (Right angle-Hypotenuse-Side) congruence rule: \[ \triangle BFC \cong \triangle CEB \] By CPCT (Corresponding Parts of Congruent Triangles): \[ \angle FBC = \angle ECB \] Therefore, the sides opposite these equal angles are also equal: \[ AB = AC \] So, \( \triangle ABC \) is an isosceles triangle.
Q5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that \(\mathrm{\angle B = \angle C.}\)
Solution:
Given:
\( \triangle ABC \) is isosceles, with \( AB = AC \).We are also given: \[ \angle APB = \angle APC = 90^{\circ} \] In \( \triangle ABP \) and \( \triangle ACP \): \[ \scriptsize \begin{aligned} AB &= AC \text{ (Given, hypotenuse for both triangles)} \\ AP &= AP \text{ (Common side)} \\ \angle APB &= \angle APC = 90^{\circ} \end{aligned} \] By RHS (Right angle-Hypotenuse-Side) Congruence Rule: \[ \triangle ABP \cong \triangle ACP \] By CPCT (Corresponding Parts of Congruent Triangles): \[ \angle B = \angle C \]
Conclusion:
Base angles \( \angle B \) and \( \angle C \) of the isosceles triangle are equal.