GSEB Class 9 Maths Solutions Chapter 7 Triangles Exercise 7.1

Question 1. In quadrilateral ABCD, \( AC = AD \), and AB bisect \( \angle A \) (see in figure). Show that \( \triangle ABC \cong \triangle ABD \). What can you say about BC and BD?
Answer:
Given: Quadrilateral ABCD in which AB is the bisector of \( \angle A \) and \( AC = AD \).
To show: \( \triangle ABC \cong \triangle ABD \)
Proof: In \( \triangle ABC \) and \( \triangle ABD \),
1. \( AC = AD \) (Given)
2. \( \angle CAB = \angle DAB \) (Since AB bisects \( \angle A \))
3. \( AB = AB \) (Common side)
Therefore, \( \triangle ABC \cong \triangle ABD \) (by SAS Congruence Rule).

What can you say about BC and BD?
Since \( \triangle ABC \cong \triangle ABD \), their corresponding parts are equal.
Thus, \( BC = BD \) (by CPCTC - Corresponding Parts of Congruent Triangles are Congruent).
In simple words: We are given a four-sided shape where two sides are equal and a line cuts an angle in half. We need to prove that two triangles inside this shape are identical. By comparing their sides and angles, we can show they match using the Side-Angle-Side rule. Because the triangles are identical, their third sides must also be equal.

Exam Tip: Remember the four main congruence rules (SSS, SAS, ASA, RHS) and CPCTC. Clearly state which sides and angles are equal and why (given, common, bisector, etc.) to get full marks.

Question 2. ABCD is a quadrilateral in which \( AD = BC \) and \( \angle DAB = \angle CBA \) (see figure). Prove that
1. \( \triangle ABD \cong \triangle BAC \)
2. \( BD = AC \)
3. \( \angle ABD = \angle BAC \)
Answer:
Given: ABCD is a quadrilateral in which \( AD = BC \) and \( \angle DAB = \angle CBA \).
Proof:
1. In \( \triangle ABD \) and \( \triangle BAC \),
  \( AD = BC \) (given)
  \( \angle DAB = \angle CBA \) (given)
  \( AB = AB \) (common side)
  \( \implies \triangle ABD \cong \triangle BAC \) (by SAS Congruence Rule)
2. Since \( \triangle ABD \cong \triangle BAC \) (proved above),
  \( \implies BD = AC \) (by CPCTC)
3. Since \( \triangle ABD \cong \triangle BAC \) (proved above),
  \( \implies \angle ABD = \angle BAC \) (by CPCTC)
In simple words: For this four-sided figure, we're told that two opposite sides and two angles are equal. First, we show that two triangles inside are identical by matching their sides and angles using the SAS rule. Because these triangles are identical, their corresponding parts, like their diagonals and certain angles, must also be equal.

Exam Tip: When proving congruence and subsequent equality of parts, always refer back to the congruence you have already established (e.g., "proved above") for each CPCTC step.

Question 3. AD and BC are equal perpendiculars to a line segment AB (see fig). Show that CD bisects AB.
Answer:
Given: AD and BC are perpendiculars on AB such that \( AD = BC \).
To Prove: CD bisects AB, meaning \( OA = OB \).
Proof: In \( \triangle AOD \) and \( \triangle BOC \),
  \( AD = BC \) (Given)
  \( \angle OAD = \angle OBC \) (Each is \( 90^\circ \))
  \( \angle ADO = \angle BCO \) (Alternate interior angles, because AD || BC since both are perpendicular to the same line AB)
  \( \implies \triangle AOD \cong \triangle BOC \) (by ASA Congruence Rule)
Therefore, \( OA = OB \) (by CPCTC)
  \( \implies \) CD bisects AB.
In simple words: We have two lines standing straight up from a base line, and they are the same length. We need to show that the line connecting their tops cuts the base line exactly in the middle. We prove two triangles are identical by comparing their angles and one side. Since the triangles are identical, the parts of the base line that they cut off must also be equal.

Exam Tip: To prove a line segment is bisected, you typically need to show that two triangles formed by the intersecting line are congruent, allowing you to use CPCTC for the equal segments.

Question 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see in figure). Show that \( \triangle ABC \cong \triangle CDA \).
Answer:
Given: l || m and p || q. These lines intersect to form quadrilateral ABCD.
To Prove: \( \triangle ABC \cong \triangle CDA \)
Proof: Since AB || DC (part of parallel lines p and q) and AD || BC (part of parallel lines l and m).
Thus, ABCD is a parallelogram.
Now in \( \triangle ABC \) and \( \triangle CDA \),
  \( \angle BAC = \angle DCA \) (Alternate interior angles, since AB || DC)
  \( AC = AC \) (Common side)
  \( \angle BCA = \angle DAC \) (Alternate interior angles, since AD || BC)
  \( \implies \triangle ABC \cong \triangle CDA \) (by ASA Congruence Rule)
In simple words: We have two sets of parallel lines that cross each other, creating a four-sided shape. We need to demonstrate that two triangles formed by these lines are identical. By using the fact that alternate interior angles between parallel lines are equal, and that a side is shared, we can prove the triangles are congruent using the Angle-Side-Angle rule.

Exam Tip: When dealing with parallel lines intersected by transversals, always look for alternate interior angles, corresponding angles, and co-interior angles, as these are key for proving congruence.

Question 5. Line l is the bisector of an angle \( \angle A \) and B is any point on l. BP and BQ are perpendiculars from B to the arms of \( \angle A \) (see figure). Show that
(i) \( \triangle APB \cong \triangle AQB \)
(ii) \( BP = BQ \) or B is equidistant from the arms of \( \angle A \).
Answer:
Given: Line l is the bisector of \( \angle A \), so \( \angle BAP = \angle BAQ \). BP and BQ are perpendiculars from B to the arms of \( \angle A \).
(i) In \( \triangle APB \) and \( \triangle AQB \),
  \( \angle BPA = \angle BQA \) (Each is \( 90^\circ \), since BP and BQ are perpendiculars)
  \( \angle BAP = \angle BAQ \) (Given, as l is the angle bisector of \( \angle A \))
  \( AB = AB \) (Common side)
  \( \implies \triangle APB \cong \triangle AQB \) (by AAS Congruence Rule)
(ii) Since \( \triangle APB \cong \triangle AQB \) (proved above),
  \( \implies BP = BQ \) (by CPCTC)
This means point B is equidistant from the arms of \( \angle A \).
In simple words: We have an angle split perfectly in half by a line. A point on this line has two other lines drawn from it, making right angles with the sides of the original angle. First, we show that the two small triangles formed are identical by looking at their angles and a common side. Because these triangles are identical, the lengths of the two lines making right angles must be equal, which means the point is the same distance from both sides of the main angle.

Exam Tip: The property that any point on the angle bisector is equidistant from the arms of the angle is a fundamental geometric theorem. Knowing this theorem helps you predict the outcome of such problems.

Question 6. In figure, \( AC = AE \), \( AB = AD \) and \( \angle BAD = \angle EAC \). Show that \( BC = DE \).
Answer:
Given: In figure, \( AC = AE \), \( AB = AD \) and \( \angle BAD = \angle EAC \).
To show: \( BC = DE \).
Proof: We are given that \( \angle BAD = \angle EAC \).
If we add \( \angle DAC \) to both sides of this equality, we get:
\( \angle BAD + \angle DAC = \angle EAC + \angle DAC \)
This simplifies to:
\( \angle BAC = \angle DAE \)
Now, consider \( \triangle ABC \) and \( \triangle ADE \).
  \( AB = AD \) (given)
  \( \angle BAC = \angle DAE \) (proved above)
  \( AC = AE \) (given)
  \( \implies \triangle ABC \cong \triangle ADE \) (by SAS Congruence Rule)
Hence, \( BC = DE \) (by CPCTC).
In simple words: We have two triangles that seem to overlap. We know that some of their sides are equal, and some parts of their angles are equal. To prove that a specific side of one triangle is equal to a specific side of the other, we first adjust the given angle information by adding a common angle part to both sides. This helps us to establish that the full angles needed for triangle congruence are equal. Then, using the Side-Angle-Side rule, we prove the two main triangles are identical, which means their remaining corresponding sides must also be equal.

Exam Tip: When angles are made up of common parts, always consider adding or subtracting the common angle to simplify the comparison and identify relevant congruent parts.

Question 7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \( \angle BAD = \angle ABE \) and \( \angle EPA = \angle DPB \) (see figure). Show that.
(i) \( \triangle DAP \cong \triangle EBP \)
(ii) \( AD = BE \)
Answer:
Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \( \angle BAD = \angle ABE \) and \( \angle EPA = \angle DPB \).
Proof:
(i) We have \( \angle EPA = \angle DPB \) (given).
Adding \( \angle DPE \) to both sides, we get:
\( \angle EPA + \angle DPE = \angle DPB + \angle DPE \)
This means \( \angle DPA = \angle EPB \).
Now, consider \( \triangle DAP \) and \( \triangle EBP \),
  \( \angle BAD = \angle ABE \) (given, also written as \( \angle DAP = \angle EBP \))
  \( AP = PB \) (Given, as P is the mid-point of AB)
  \( \angle DPA = \angle EPB \) (proved above)
  \( \implies \triangle DAP \cong \triangle EBP \) (by ASA Congruence Rule)
(ii) Since \( \triangle DAP \cong \triangle EBP \) (proved above),
  \( \implies AD = BE \) (by CPCTC).
In simple words: We have a line segment with a middle point, and two points above it. We're given that certain angles are equal. First, we need to adjust one of the angle equalities by adding a shared angle part, so we have the full angles needed. Then, by using the Angle-Side-Angle rule, we show that two triangles are identical, since they share a middle point on the base. Because these triangles are identical, their matching sides must also be equal.

Exam Tip: When angles are adjacent and share a common ray, carefully analyze how adding or subtracting a common angle can help in establishing equality for the entire angle needed for congruence proofs.

Question 8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that \( DM = CM \). Point D is joined to point B (see figure). Show that:
1. \( \triangle AMC \cong \triangle BMD \)
2. \( \angle DBC \) is a right angle.
3. \( \triangle DBC \cong \triangle ACB \)
4. \( CM = \frac{1}{2}AB \)
Answer:
Given: Right triangle ABC, \( \angle C = 90^\circ \). M is the mid-point of hypotenuse AB. C is joined to M and produced to D such that \( DM = CM \). D is joined to B.
Proof:
1. In \( \triangle AMC \) and \( \triangle BMD \),
  \( AM = BM \) (Given, as M is the mid-point of AB)
  \( \angle AMC = \angle BMD \) (Vertically opposite angles)
  \( CM = DM \) (Given)
  \( \implies \triangle AMC \cong \triangle BMD \) (by SAS Congruence Rule)
2. Since \( \triangle AMC \cong \triangle BMD \) (Proved in part 1),
  \( \implies AC = BD \) (by CPCTC)
  And \( \angle ACM = \angle BDM \) (by CPCTC)
These are alternate interior angles. Since they are equal, AC must be parallel to BD (AC || BD).
When two parallel lines are cut by a transversal, the sum of consecutive interior angles is \( 180^\circ \).
So, \( \angle ACB + \angle DBC = 180^\circ \).
Since \( \angle ACB = 90^\circ \) (given),
  \( 90^\circ + \angle DBC = 180^\circ \)
  \( \implies \angle DBC = 180^\circ - 90^\circ \)
  \( \implies \angle DBC = 90^\circ \).
Thus, \( \angle DBC \) is a right angle.
3. Now, consider \( \triangle DBC \) and \( \triangle ACB \),
  \( BD = AC \) (Proved in part 2)
  \( \angle DBC = \angle ACB \) (Each is \( 90^\circ \), proved in part 2 and given)
  \( BC = CB \) (Common side)
  \( \implies \triangle DBC \cong \triangle ACB \) (by SAS Congruence Rule)
4. Since \( \triangle DBC \cong \triangle ACB \) (Proved in part 3),
  \( \implies DC = AB \) (by CPCTC)
We know that \( DC = DM + CM \). Since \( DM = CM \) (given),
  \( DC = CM + CM = 2CM \)
So, \( 2CM = AB \)
  \( \implies CM = \frac{1}{2}AB \)
In simple words: We have a right-angled triangle. The middle point of its longest side is extended to create another point, making two segments equal. Then, new lines are drawn. We need to prove four things: First, two small triangles are identical using the SAS rule. Second, a specific angle in the new figure is a right angle because parallel lines have interior angles that add up to 180 degrees. Third, two larger triangles are identical, again using SAS. Finally, we prove that the line segment from the right angle to the midpoint of the hypotenuse is exactly half the length of the hypotenuse, using the properties of congruent triangles.

Exam Tip: This problem connects triangle congruence with properties of parallel lines and midpoints. Remember that the converse of the Alternate Interior Angles Theorem is key for showing lines are parallel, and CPCTC is fundamental for deriving lengths and angles.