CBSE Class 9 Mathematics Chapter 7: Triangles NCERT Solutions

To prove that $\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; ABD$, we need to show that the corresponding sides and angles of the two triangles are equal, using one of the congruence criteria.

Consider $\backslash triangle\; ABC$ and $\backslash triangle\; ABD$. We are given the following information:

1. $AC\; =\; AD$ (Given)
2. $\backslash angle\; CAB\; =\; \backslash angle\; DAB$ (Since AB bisects $\backslash angle\; A$)
3. $AB\; =\; AB$ (Common side to both triangles)

By the SAS (Side-Angle-Side) congruence criterion, since two sides and the included angle of $\backslash triangle\; ABC$ are equal to the corresponding two sides and the included angle of $\backslash triangle\; ABD$, we can conclude that:

$\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; ABD$

Now, we need to determine the relationship between BC and BD. Since the triangles are congruent, their corresponding parts are equal (CPCT - Corresponding Parts of Congruent Triangles).

Therefore, $BC\; =\; BD$.

Answer: $BC\; =\; BD$.

We are given a quadrilateral ABCD with $AD\; =\; BC$ and $\backslash angle\; DAB\; =\; \backslash angle\; CBA$. We need to prove the given statements.

(i) To prove $\backslash triangle\; ABD\; \backslash cong\; \backslash triangle\; BAC$:

Consider $\backslash triangle\; ABD$ and $\backslash triangle\; BAC$. We have:

1. $AD\; =\; BC$ (Given)
2. $\backslash angle\; DAB\; =\; \backslash angle\; CBA$ (Given)
3. $AB\; =\; AB$ (Common side)

By the SAS (Side-Angle-Side) congruence criterion, since two sides and the included angle of $\backslash triangle\; ABD$ are equal to the corresponding two sides and the included angle of $\backslash triangle\; BAC$, we have:

$\backslash triangle\; ABD\; \backslash cong\; \backslash triangle\; BAC$

(ii) To prove $BD\; =\; AC$:

Since $\backslash triangle\; ABD\; \backslash cong\; \backslash triangle\; BAC$ (proved in part i), their corresponding parts are equal (CPCT).

Therefore, $BD\; =\; AC$.

(iii) To prove $\backslash angle\; ABD\; =\; \backslash angle\; BAC$:

Again, since $\backslash triangle\; ABD\; \backslash cong\; \backslash triangle\; BAC$ (proved in part i), their corresponding parts are equal (CPCT).

Therefore, $\backslash angle\; ABD\; =\; \backslash angle\; BAC$.

Proved.

We are given that AD and BC are equal perpendiculars to the line segment AB. This means $AD\; \backslash perp\; AB$, $BC\; \backslash perp\; AB$, and $AD\; =\; BC$. We need to show that CD bisects AB, which means we need to prove that the point of intersection of CD and AB, let's call it O, is the midpoint of AB (i.e., $AO\; =\; BO$).

Consider the triangles $\backslash triangle\; AOD$ and $\backslash triangle\; BOC$.

We have the following information:

1. $\backslash angle\; DAO\; =\; \backslash angle\; CBO\; =\; 90^\{\backslash circ\}$ (Since AD and BC are perpendiculars to AB)
2. $AD\; =\; BC$ (Given)
3. $\backslash angle\; AOD\; =\; \backslash angle\; BOC$ (Vertically opposite angles are equal)

By the AAS (Angle-Angle-Side) congruence criterion, since two angles and a non-included side of $\backslash triangle\; AOD$ are equal to the corresponding two angles and the non-included side of $\backslash triangle\; BOC$, we can conclude that:

$\backslash triangle\; AOD\; \backslash cong\; \backslash triangle\; BOC$

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Therefore, $AO\; =\; BO$.

This means that the point O, which is the intersection of CD and AB, is the midpoint of AB. Hence, CD bisects AB.

Proved.

We are given two pairs of parallel lines, l || m and p || q. Let the intersection points form a quadrilateral ABCD. Line AC is a diagonal. We need to show that $\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; CDA$.

Since p || q, we can consider AC as a transversal. Therefore, the alternate interior angles are equal:

$\backslash angle\; BAC\; =\; \backslash angle\; DCA$ (Alternate angles)

Since l || m, we can again consider AC as a transversal. Therefore, the alternate interior angles are equal:

$\backslash angle\; BCA\; =\; \backslash angle\; DAC$ (Alternate angles)

Now, consider the triangles $\backslash triangle\; ABC$ and $\backslash triangle\; CDA$.

We have:

1. $\backslash angle\; BAC\; =\; \backslash angle\; DCA$ (Proved above)
2. $AC\; =\; AC$ (Common side)
3. $\backslash angle\; BCA\; =\; \backslash angle\; DAC$ (Proved above)

By the ASA (Angle-Side-Angle) congruence criterion, since two angles and the included side of $\backslash triangle\; ABC$ are equal to the corresponding two angles and the included side of $\backslash triangle\; CDA$, we can conclude that:

$\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; CDA$

Proved.

We are given that line l bisects $\backslash angle\; A$, and B is a point on l. BP and BQ are perpendiculars from B to the arms of $\backslash angle\; A$. This means $\backslash angle\; PAB\; =\; \backslash angle\; QAB$ and $\backslash angle\; APB\; =\; \backslash angle\; AQB\; =\; 90^\{\backslash circ\}$.

(i) To prove $\backslash triangle\; APB\; \backslash cong\; \backslash triangle\; AQB$:

Consider $\backslash triangle\; APB$ and $\backslash triangle\; AQB$. We have:

1. $\backslash angle\; PAB\; =\; \backslash angle\; QAB$ (Since l bisects $\backslash angle\; A$)
2. $\backslash angle\; APB\; =\; \backslash angle\; AQB\; =\; 90^\{\backslash circ\}$ (Given that BP and BQ are perpendiculars)
3. $AB\; =\; AB$ (Common side)

By the AAS (Angle-Angle-Side) congruence criterion, since two angles and a non-included side of $\backslash triangle\; APB$ are equal to the corresponding two angles and the non-included side of $\backslash triangle\; AQB$, we have:

$\backslash triangle\; APB\; \backslash cong\; \backslash triangle\; AQB$

(ii) To prove $BP\; =\; BQ$:

Since $\backslash triangle\; APB\; \backslash cong\; \backslash triangle\; AQB$ (proved in part i), their corresponding parts are equal (CPCT).

Therefore, $BP\; =\; BQ$.

This shows that point B is equidistant from the arms of $\backslash angle\; A$.

Proved.

We are given $AC\; =\; AE$, $AB\; =\; AD$, and $\backslash angle\; BAD\; =\; \backslash angle\; EAC$. We need to show that $BC\; =\; DE$.

First, let's establish the relationship between the angles $\backslash angle\; BAC$ and $\backslash angle\; DAE$.

We are given: $\backslash angle\; BAD\; =\; \backslash angle\; EAC$

Add $\backslash angle\; DAC$ to both sides of the equation:

$\backslash angle\; BAD\; +\; \backslash angle\; DAC\; =\; \backslash angle\; EAC\; +\; \backslash angle\; DAC$

Observing the figure, we can see that $\backslash angle\; BAD\; +\; \backslash angle\; DAC$ forms $\backslash angle\; BAC$, and $\backslash angle\; EAC\; +\; \backslash angle\; DAC$ forms $\backslash angle\; DAE$.

Therefore, $\backslash angle\; BAC\; =\; \backslash angle\; DAE$ ... (i)

Now, consider the triangles $\backslash triangle\; ABC$ and $\backslash triangle\; ADE$.

We have the following information:

1. $AB\; =\; AD$ (Given)
2. $AC\; =\; AE$ (Given)
3. $\backslash angle\; BAC\; =\; \backslash angle\; DAE$ (From equation (i))

By the SAS (Side-Angle-Side) congruence criterion, since two sides and the included angle of $\backslash triangle\; ABC$ are equal to the corresponding two sides and the included angle of $\backslash triangle\; ADE$, we can conclude that:

$\backslash triangle\; ABC\; \backslash cong\; \backslash triangle\; ADE$

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Therefore, $BC\; =\; DE$.

Proved.