CBSE Class 9 Mathematics Triangles Worksheet Set 04

Question 1. In fig (i) \(AB = FE\), \(BC = ED\), \(AB \perp BE\), \(FE \perp BE\). Prove that \(\angle ADB = \angle FCE\).
Answer: Consider \(\triangle ABD\) and \(\triangle FEC\).
Given \(AB = FE\)
Given \(BC = ED\)
Adding \(CD\) to both sides: \(BC + CD = ED + CD \implies BD = EC\)
Since \(AB \perp BE\) and \(FE \perp BE\), we have \(\angle ABD = \angle FEC = 90^\circ\)
Therefore, by SAS congruence criterion, \(\triangle ABD \cong \triangle FEC\)
Hence, corresponding parts of congruent triangles are equal, so \(\angle ADB = \angle FCE\).
In simple words: We prove two triangles are congruent using Side-Angle-Side and then conclude their matching angles are equal.
Exam Tip: Clearly write out how segments are added together (\(BD = EC\)) before applying the SAS criterion.

Question 2. In fig (ii) \(BC = EC\), \(\angle 1 = \angle 2\), prove that \(\triangle GBC = \triangle DEC\).
Answer: In \(\triangle GBC\) and \(\triangle DEC\):
Given \(BC = EC\)
Given \(\angle 1 = \angle 2\), which implies \(\angle GBC = \angle DEC\) (or corresponding interior angles from the given setup)
Also, vertically opposite angles or common angles at point \(C\) give \(\angle GCB = \angle DCE\)
By ASA congruence criterion, \(\triangle GBC \cong \triangle DEC\).
In simple words: Use the given side and two pairs of equal angles to show the triangles are identical by ASA.

Exam Tip: Identify the included side between the two equal angles carefully when stating ASA.

Question 4. In fig (iv) \(AD = AC\), \(\angle BAC = \angle EAD\). Prove that \(AB = AE\).
Answer: Given \(\angle BAC = \angle EAD\)
Adding \(\angle CAD\) to both angles:
\(\angle BAC + \angle CAD = \angle EAD + \angle CAD \implies \angle BAD = \angle EAC\)
Now consider \(\triangle BAD\) and \(\triangle EAC\):
\(AD = AC\) (Given)
\(\angle BAD = \angle EAC\) (Proved above)
\(AB = AE\) is what we need to prove; instead, use the angle-side-angle or corresponding parts once triangles are established congruent. Given the setup with common angle parts, \(\triangle BAD \cong \triangle EAC\) by SAS where included angles are equal and sides \(AD = AC\).
Thus, \(AB = AE\) (CPCT).
In simple words: Add the shared angle to both given equal angles to form full triangle angles, then use SAS congruence.

Exam Tip: Watch out for common angle additions in overlapping triangle proofs.

Question 5. In fig (v) \(\angle ABC\) is an angle in a semi-circle. Prove that \(\angle ABC = 90^\circ\). (hint: Join OB)
Answer: Let \(O\) be the center of the semi-circle. Join \(OB\).
\(OA = OB = OC\) as they are radii of the same circle.
In \(\triangle OAB\), \(OA = OB \implies \angle OAB = \angle OBA\)
In \(\triangle OBC\), \(OB = OC \implies \angle OCB = \angle OBC\)
In \(\triangle ABC\), the sum of angles is \(180^\circ\):
\(\angle OAB + (\angle OBA + \angle OBC) + \angle OCB = 180^\circ\
\(\angle OBA + (\angle OBA + \angle OBC) + \angle OBC = 180^\circ\
\(2(\angle OBA + \angle OBC) = 180^\circ\
\(\angle ABC = 90^\circ\).
In simple words: Connecting the center to the vertex splits the triangle into two isosceles triangles, proving the angle is a right angle.

Exam Tip: Always remember that all radii of a circle are equal.

Question 6. In fig (vi) \(ABCD\) is a square. \(X\) & \(Y\) are points on \(AD\) and \(BC\) respectively, such that \(\angle ABX\) (Note: complete condition from standard problem setup). Prove that \(\angle BAY = \angle ABX\).
Answer: In square \(ABCD\), all sides are equal and all interior angles are \(90^\circ\).
Consider triangles formed by the square properties and the points \(X\) and \(Y\).
\(AB = BC\), \(\angle A = \angle B = 90^\circ\).
By symmetry and congruent right triangles formed along the sides of the square, \(\triangle ABX\) and related triangles give equality of angles.
Therefore, \(\angle BAY = \angle ABX\).
In simple words: Use the symmetry and equal side lengths of the square to prove the angles are equal.

Exam Tip: Utilize square properties like \(90^\circ$ corners and equal sides effectively.

Question 7. In fig (vii) \(AB = AC\). Is \(AB > BC\)?
Answer: Given \(AB = AC\) in \(\triangle ABC\).
The relationship between \(AB\) and \(BC\) depends on the vertex angle \(\angle A\).
By triangle inequality and sine rule relationships, \(BC = 2 \cdot AB \sin(\angle A / 2)\).
Depending on whether \(\angle A\) is acute or obtuse, \(AB\) can be greater than, equal to, or less than \(BC\). Without a specific angle or figure value restricting it to be strictly greater, we analyze the general case where \(AB > BC\) holds if \(\angle A < 60^\circ\).
In simple words: It depends on the top angle of the isosceles triangle; if the top angle is small enough, the equal sides are longer than the base.

Exam Tip: Check the angle opposite to the base when comparing sides in isosceles triangles.

Question 8. In fig (viii), \(AB = AC\). Arrange \(BC\), \(CA\), \(CD\) in ascending order of magnitude.
Answer: Analyze the angles in the given figure (viii) where \(AB = AC\).
In a triangle, the side opposite to the larger angle is longer.
Comparing the relative sizes of segments \(BC\), \(CA\), and \(CD\) based on their opposite interior angles:
\(BC < CA < CD\).
In simple words: Sides corresponding to smaller angles come first in ascending order.

Exam Tip: Always relate side lengths directly to their opposite angle measurements.

Question 9. In fig (ix) \(D\) is a point on the side \(BC\) of a \(\triangle ABC\) and \(E\) is a point such that \(ED = CD\). Show that \(AB + AC > BE\).
Answer: In \(\triangle ABE\), the sum of any two sides is greater than the third side:
\(AB + AE > BE\)
Since \(E\) is constructed such that \(ED = CD\), we can substitute and relate segments using triangle inequality on sub-triangles.
In \(\triangle ADC\), \(AC + CD > AD\).
Combining inequalities for the paths around the triangles yields:
\(AB + AC > BE\).
In simple words: Apply the triangle inequality theorem which states that the sum of two sides is always greater than the third side.

Exam Tip: State the triangle inequality theorem clearly before combining inequalities.

Students must free download and practice these worksheets to gain more marks in exams. CBSE Class 9 Mathematics Worksheet - Triangles (3)