Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 3 Set 3.1 Triangles Solutions

Question 1. In the adjoining figure, \(\angle ACD\) is an exterior angle of \(\triangle ABC\). \(\angle B = 40^\circ\), \(\angle A = 70^\circ\). Find the measure of \(\angle ACD\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें B, C, D एक सीधी रेखा पर हैं। कोण ACD त्रिभुज ABC का एक बाहरी कोण है। शीर्ष A, B, C और D को स्पष्ट रूप से लेबल किया गया है।
Answer:
Solution:
\(\angle A = 70^\circ\), \(\angle B = 40^\circ\) [Given]
\(\angle ACD\) is an exterior angle of \(\triangle ABC\). [Given]
\(\therefore \angle ACD = \angle A + \angle B\)
\(= 70^\circ + 40^\circ\)
\(\therefore \angle ACD = 110^\circ\)
In simple words: The exterior angle of a triangle is equal to the sum of its two opposite interior angles. By adding the given angles A and B, we find the measure of the exterior angle ACD.

🎯 Exam Tip: Remember the exterior angle theorem, which states that the measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles. This is a fundamental concept for scoring well in geometry problems.

Question 2. In \(\triangle PQR\), \(\angle P = 70^\circ\), \(\angle Q = 65^\circ\), then find \(\angle R\).
Answer:
Solution:
\(\angle P = 70^\circ\), \(\angle Q = 65^\circ\) [Given]
In \(\triangle PQR\),
\(\angle P + \angle Q + \angle R = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 70^\circ + 65^\circ + \angle R = 180^\circ\)
\(\therefore \angle R = 180^\circ - 70^\circ - 65^\circ\)
\(\therefore \angle R = 45^\circ\)
In simple words: The sum of all interior angles in any triangle is \(180^\circ\). Given two angles, we subtract their sum from \(180^\circ\) to find the third angle.

🎯 Exam Tip: Always state the property being used (e.g., "Sum of angles in a triangle is \(180^\circ\)") as a reason in bracket to earn full marks. Accuracy in calculation is key.

Question 3. The measures of angles of a triangle are \(x^\circ\), \((x - 20)^\circ\), \((x - 40)^\circ\). Find the measure of each angle.
Answer:
Solution:
The measures of the angles of a triangle are \(x^\circ\), \((x - 20)^\circ\), \((x - 40)^\circ\). [Given]
\(\therefore x^\circ + (x - 20)^\circ + (x - 40)^\circ = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 3x - 60 = 180\)
\(\therefore 3x = 180 + 60\)
\(\therefore 3x = 240\)
\(\therefore x = \frac{240}{3}\)
\(\therefore x = 80^\circ\)
\(\therefore\) The measures of the remaining angles are
\(x - 20^\circ = 80^\circ - 20^\circ = 60^\circ\),
\(x - 40^\circ = 80^\circ - 40^\circ = 40^\circ\)
\(\therefore\) The measures of the angles of the triangle are \(80^\circ\), \(60^\circ\) and \(40^\circ\).
In simple words: We use the property that the sum of angles in a triangle is \(180^\circ\). By setting up an equation with the given algebraic expressions for the angles, we solve for 'x' and then find each angle's measure.

🎯 Exam Tip: Ensure that after finding the value of 'x', you substitute it back into all the angle expressions to find the measure of each individual angle. Double-check your arithmetic.

Question 4. The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.
Answer:
Solution:
Let the measure of the smallest angle be \(x^\circ\).
One of the angles is twice the measure of the smallest angle.
\(\therefore\) Measure of that angle = \(2x^\circ\)
Another angle is thrice the measure of the smallest angle.
\(\therefore\) Measure of that angle = \(3x^\circ\)
\(\therefore\) The measures of the remaining two angles are \(2x^\circ\) and \(3x^\circ\).
Now, \(x^\circ + 2x^\circ + 3x^\circ = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 6x = 180\)
\(\therefore x = \frac{180}{6}\)
\(\therefore x^\circ = 30^\circ\)
The measures of the remaining angles are \(2x^\circ = 2 \times 30^\circ = 60^\circ\)
\(3x^\circ = 3 \times 30^\circ = 90^\circ\)
The measures of the three angles of the triangle are \(30^\circ\), \(60^\circ\) and \(90^\circ\).
In simple words: We represent the smallest angle as 'x' and express the other two angles in terms of 'x'. Then, using the angle sum property of a triangle, we solve for 'x' and calculate each angle's measure.

🎯 Exam Tip: Clearly define your variables (e.g., "Let the smallest angle be \(x^\circ\)") at the beginning of the solution. This clarity helps in setting up the equation correctly and avoids errors.

Question 5. In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ENM को दर्शाता है, जिसमें T, E, N, M और R बिंदु हैं। कोण NET \(100^\circ\) है, और कोण EMR \(140^\circ\) है। कोण ENM को x, कोण NEM को y, और कोण EMN को z से चिह्नित किया गया है।
Answer:
Solution:
(i) \(\angle NET = 100^\circ\) and \(\angle EMR = 140^\circ\)
\(\angle EMN + \angle EMR = 180^\circ\) (Angles in a linear pair)
\(\therefore z + 140^\circ = 180^\circ\)
\(\therefore z = 180^\circ - 140^\circ\)
\(\therefore z = 40^\circ\)
(ii) Also, \(\angle NET + \angle NEM = 180^\circ\) [Angles in a linear pair]
\(\therefore 100^\circ + y = 180^\circ\)
\(\therefore y = 180^\circ - 100^\circ\)
\(\therefore y = 80^\circ\)
(iii) In \(\triangle ENM\),
\(\therefore \angle ENM + \angle NEM + \angle EMN = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore x + 80^\circ + 40^\circ = 180^\circ\)
\(\therefore x = 180^\circ - 80^\circ - 40^\circ\)
\(\therefore x = 60^\circ\)
\(\therefore x = 60^\circ\), \(y = 80^\circ\), \(z = 40^\circ\)
In simple words: We find 'z' and 'y' using the property of angles in a linear pair. Then, using the sum of angles in a triangle, we find 'x'.

🎯 Exam Tip: When dealing with multiple angles and variables, break down the problem into smaller steps. Clearly identify which geometric property (e.g., linear pair, angle sum property) is used for each calculation. This systematic approach ensures accuracy.

Question 6. In the adjoining figure, line AB || line DE. Find the measures of \(\angle DRE\) and \(\angle ARE\) using given measures of some angles.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाएँ AB और DE को दर्शाता है, जिन्हें एक तिर्यक रेखा AD काट रही है। बिंदु R, AD और BE के प्रतिच्छेदन पर है। कोण BAD \(70^\circ\) और कोण DER \(40^\circ\) के रूप में दिए गए हैं।
Answer:
Solution:
(i) \(\angle BAD = 70^\circ\), \(\angle DER = 40^\circ\) [Given]
line AB || line DE and seg AD is their transversal.
\(\therefore \angle EDA = \angle BAD\) [Alternate Angles]
\(\therefore \angle EDA = 70^\circ\) ....(i)
In \(\triangle DRE\),
\(\angle EDR + \angle DER + \angle DRE = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 70^\circ + 40^\circ + \angle DRE = 180^\circ\) [From (i) and D - R - A]
\(\therefore \angle DRE = 180^\circ - 70^\circ - 40^\circ\)
\(\therefore \angle DRE = 70^\circ\)
(ii) \(\angle DRE + \angle ARE = 180^\circ\) [Angles in a linear pair]
\(\therefore 70^\circ + \angle ARE = 180^\circ\)
\(\therefore \angle ARE = 180^\circ - 70^\circ\)
\(\therefore \angle ARE = 110^\circ\)
\(\therefore \angle DRE = 70^\circ\), \(\angle ARE = 110^\circ\)
In simple words: First, we use the alternate angles property to find \(\angle EDA\). Then, using the angle sum property of \(\triangle DRE\), we find \(\angle DRE\). Finally, we use the linear pair property to find \(\angle ARE\).

🎯 Exam Tip: Clearly identify parallel lines and transversals to correctly apply properties like alternate angles and corresponding angles. Always provide reasons for each step to validate your geometric deductions.

Question 7. In \(\triangle ABC\), bisectors of \(\angle A\) and \(\angle B\) intersect at point O. If \(\angle C = 70^\circ\), find the measure of \(\angle AOB\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें कोण A और कोण B के समद्विभाजक बिंदु O पर प्रतिच्छेद करते हैं। कोण ACB \(70^\circ\) दिया गया है।
Answer:
Solution:
\(\angle OAB = \angle OAC = \frac{1}{2} \angle BAC\) ....(i) [Seg AO bisects \(\angle BAC\)]
\(\angle OBA = \angle OBC = \frac{1}{2} \angle ABC\) .....(ii) [Seg RO bisects \(\angle ABC\)]
In \(\triangle ABC\),
\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore \angle BAC + \angle ABC + 70^\circ = 180^\circ\)
\(\therefore \angle BAC + \angle ABC = 180^\circ - 70^\circ\)
\(\therefore \angle BAC + \angle ABC = 110^\circ\)
\(\frac{1}{2} (\angle BAC) + \frac{1}{2} (\angle ABC) = \frac{1}{2} \times 110^\circ\) [Multiplying both sides by \(\frac{1}{2}\)]
\(\therefore \angle OAB + \angle OBA = 55^\circ\) ....(iii) [From (i) and (ii)]
In \(\triangle OAB\),
\(\angle OAB + \angle OBA + \angle AOB = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 55^\circ + \angle AOB = 180^\circ\) [From (iii)]
\(\therefore \angle AOB = 180^\circ - 55^\circ\)
\(\therefore \angle AOB = 125^\circ\)
In simple words: First, find the sum of \(\angle BAC\) and \(\angle ABC\) using the angle sum property of \(\triangle ABC\). Since AO and BO are angle bisectors, \(\angle OAB\) and \(\angle OBA\) are half of \(\angle BAC\) and \(\angle ABC\) respectively. Use this to find the sum \(\angle OAB + \angle OBA\). Finally, apply the angle sum property to \(\triangle OAB\) to find \(\angle AOB\).

🎯 Exam Tip: When angle bisectors are involved, remember that they divide the angle into two equal parts. This is crucial for setting up the correct relationships between angles. Carefully track your steps and calculations, especially when dealing with fractions.

Question 8. In the adjoining figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of \(\angle BPQ\) and \(\angle PQD\) respectively. Prove that \(m \angle PTQ = 90^\circ\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाएँ AB और CD को दर्शाता है, जिन्हें एक तिर्यक रेखा PQ काट रही है। किरण PT कोण BPQ की समद्विभाजक है और किरण QT कोण PQD की समद्विभाजक है। बिंदु T, PT और QT के प्रतिच्छेदन पर है।
Answer:
Solution:
Proof:
\(\angle TPB = \angle TPQ = \frac{1}{2} \angle BPQ\) ....(i) [Ray PT bisects \(\angle BPQ\)]
\(\angle TQD = \angle TQP = \frac{1}{2} \angle PQD\) ....(ii) [Ray QT bisects \(\angle PQD\)]
line AB || line CD and line PQ is their transversal. [Given]
\(\therefore \angle BPQ + \angle PQD = 180^\circ\) [Interior angles]
\(\therefore \frac{1}{2} (\angle BPQ) + \frac{1}{2} (\angle PQD) = \frac{1}{2} \times 180^\circ\) [Multiplying both sides by \(\frac{1}{2}\)]
\(\therefore \angle TPQ + \angle TQP = 90^\circ\)
In \(\triangle PTQ\),
\(\angle TPQ + \angle TQP + \angle PTQ = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 90^\circ + \angle PTQ = 180^\circ\) [From (iii)]
\(\therefore \angle PTQ = 180^\circ - 90^\circ\)
\(= 90^\circ\)
\(\therefore m\angle PTQ = 90^\circ\)
In simple words: We use the properties of parallel lines to show that the sum of \(\angle BPQ\) and \(\angle PQD\) is \(180^\circ\). Since PT and QT are angle bisectors, the sum of \(\angle TPQ\) and \(\angle TQP\) is \(90^\circ\). Then, applying the angle sum property in \(\triangle PTQ\), we prove \(\angle PTQ = 90^\circ\).

🎯 Exam Tip: This is a common proof problem. Ensure you clearly state the reasons for each step, especially when using properties of parallel lines (interior angles) and angle bisectors. The logical flow of the proof is critical.

Question 9. Using the information in the adjoining figure, find the measures of \(\angle a\), \(\angle b\) and \(\angle c\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज को दर्शाता है, जिसमें एक शीर्ष पर \(70^\circ\) और दूसरे शीर्ष पर \(100^\circ\) का कोण दिया गया है। आंतरिक कोणों को a, b, c के रूप में चिह्नित किया गया है। कोण \(70^\circ\) शीर्ष पर एक रैखिक युग्म बनाता है, और कोण \(100^\circ\) दूसरे शीर्ष पर एक रैखिक युग्म बनाता है।
Answer:
Solution:
(i) \(\angle c + 100^\circ = 180^\circ\) [Angles in a linear pair]
\(\therefore \angle c = 180^\circ - 100^\circ\)
\(\therefore \angle c = 80^\circ\)
(ii) \(\angle b = 70^\circ\) [Vertically opposite angles]
(iii) \(\angle a + \angle b + \angle c = 180^\circ\) [Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\angle a + 70^\circ + 80^\circ = 180^\circ\)
\(\therefore \angle a = 180^\circ - 70^\circ - 80^\circ\)
\(\therefore \angle a = 30^\circ\)
\(\therefore \angle a = 30^\circ\), \(\angle b = 70^\circ\), \(\angle c = 80^\circ\)
In simple words: We find \(\angle c\) using the linear pair property. Then, we find \(\angle b\) using the vertically opposite angles property. Finally, we use the angle sum property of the triangle to find \(\angle a\).

🎯 Exam Tip: Recognize and correctly apply different angle properties (linear pair, vertically opposite angles, angle sum property) in sequence. Labelling the angles clearly on your diagram (if allowed) can help avoid confusion.

Question 10. In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of \(\angle DEF\) and \(\angle DFM\) respectively. Prove that,
(i) \(\angle DEG = \frac{1}{2} \angle EDF\)
(ii) EF = FG
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाएँ DE और GF को दर्शाता है। बिंदु E से किरण EG निकलती है जो कोण DEF को समद्विभाजित करती है, और बिंदु F से किरण FG निकलती है जो कोण DFM को समद्विभाजित करती है। कोण DEG को \(x^\circ\) और कोण GFM को \(y^\circ\) से चिह्नित किया गया है।
Answer:
Solution:
(i)
\(\angle DEG = \angle FEG = x^\circ\) ....(i) [Ray EG bisects \(\angle DEF\)]
\(\angle GFD = \angle GFM = y^\circ\) .....(ii) [Ray FG bisects \(\angle DFM\)]
line DE || line GF and DF is their transversal. [Given]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाएँ DE और GF को दर्शाता है, जिसमें DF एक तिर्यक रेखा है। कोण EDF और कोण GFD को वैकल्पिक आंतरिक कोणों के रूप में दिखाया गया है, जहां कोण GFD को \(y^\circ\) से चिह्नित किया गया है।
\(\therefore \angle EDF = \angle GFD\) [Alternate angles]
\(\therefore \angle EDF = y^\circ\) ....(iii) [From (ii)]
line DE || line GF and EM is their transversal. [Given]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाएँ DE और GF को दर्शाता है, जिसमें EM एक तिर्यक रेखा है। कोण DEF और कोण GFM को संगत कोणों के रूप में दिखाया गया है, जहां कोण DEG को \(x^\circ\) से चिह्नित किया गया है।
\(\therefore \angle DEF = \angle GFM\) [Corresponding angles]
\(\therefore \angle DEG + \angle FEG = \angle GFM\) [Angle addition property]
\(\therefore x^\circ + x^\circ = y^\circ\) [From (i) and (ii)]
\(\therefore 2x^\circ = y^\circ\)
\(\therefore x^\circ = \frac{1}{2} y^\circ\)
\(\therefore \angle DEG = \frac{1}{2} \angle EDF\) [From (i) and (iii)]
(ii) line DE || line GF and GE is their transversal. [Given]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो समांतर रेखाएँ DE और GF को दर्शाता है, जिसमें GE एक तिर्यक रेखा है। कोण DEG और कोण FGE को वैकल्पिक आंतरिक कोणों के रूप में दिखाया गया है।
\(\therefore \angle DEG = \angle FGE\) ....(iv) [Alternate angles]
\(\therefore \angle FEG = \angle FGE\) ....(v) [From (i) and (iv)]
\(\therefore\) In \(\triangle FEG\),
\(\angle FEG = \angle FGE\) [From (v)]
\(\therefore EF = FG\) [Converse of isosceles triangle theorem]
In simple words: For part (i), we use alternate angles and angle bisector properties to show that \(\angle EDF = y^\circ\) and \(\angle DEF = 2x^\circ\). Since DE || GF, \(\angle DEF = \angle GFM = y^\circ\), leading to \(2x^\circ = y^\circ\), which implies \(\angle DEG = \frac{1}{2} \angle EDF\). For part (ii), we use alternate angles to show \(\angle DEG = \angle FGE\). Since \(\angle DEG = \angle FEG\) (due to EG being an angle bisector), it implies \(\angle FEG = \angle FGE\), making \(\triangle FEG\) an isosceles triangle with EF = FG.

🎯 Exam Tip: This proof combines several geometric concepts: parallel lines (alternate/corresponding angles), angle bisectors, and properties of isosceles triangles. Clearly mark all relationships in your diagram and meticulously justify each step for a complete and accurate proof.

Maharashtra Board Class 9 Chapter 3 Triangles Practice Set 3.1 Intext Questions And Activities

Question 1. Can you give an alternative proof of the above theorem by drawing a line through point R and parallel to seg PQ in the above figure? (Textbook pg. no. 25)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज PQS को दर्शाता है जिसमें बिंदु R, QS पर स्थित है। एक बिंदु M भी दिखाया गया है। यह त्रिभुज के बाह्य कोण प्रमेय से संबंधित हो सकता है, जहाँ बिंदु R से PQ के समानांतर एक रेखा खींचने पर विचार किया जा रहा है।
Answer:
Solution:
Yes.
Construction: Draw line RM parallel to seg PQ through a point R.
Proof:
seg PQ || line RM and seg PR is their transversal. [Construction]
\(\therefore \angle PRM = \angle QPR\) ........(i) [Alternate angles]
seg PQ || line RM and seg QR is their transversal. [Construction]
\(\therefore \angle SRM = \angle PQR\) ........(ii) [Corresponding angles]
\(\therefore \angle PRM + \angle SRM = \angle QPR + \angle PQR\) [Adding (i) and (ii)]
\(\therefore \angle PRS = \angle PQR + \angle QPR\) [Angle addition property]
In simple words: Yes, an alternative proof can be given. By drawing a line RM parallel to PQ, we can use alternate angles and corresponding angles to show that the exterior angle \(\angle PRS\) is equal to the sum of the remote interior angles \(\angle PQR\) and \(\angle QPR\).

🎯 Exam Tip: Construction-based proofs often involve drawing auxiliary lines. Ensure your construction is valid and clearly explained. The properties of parallel lines are fundamental in such proofs, so cite them accurately.

Question 2. Observe the given figure and find the measures of \(\angle PRS\) and \(\angle RTS\). (Textbook pg. no.25)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज PRS को दर्शाता है जिसमें बिंदु T, RS पर स्थित है। कोण QPR \(40^\circ\) है, कोण PQR \(30^\circ\) है, और कोण PSR \(20^\circ\) है।
Answer:
Solution:
\(\angle PRS\) is an exterior angle of \(\triangle PQR\).
So from the theorem of remote interior angles,
\(\angle PRS = \angle PQR + \angle QPR\)
\(= 40^\circ + 30^\circ\)
\(\therefore \angle PRS = 70^\circ\)
\(\therefore \angle TRS = 70^\circ\) ... [P - T - R]
In \(\triangle RTS\),
\(\angle TRS + \angle RTS + \angle TSR = 180^\circ\) ...[Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore 70^\circ + \angle RTS + 20^\circ = 180^\circ\)
\(\therefore \angle RTS + 90^\circ = 180^\circ\)
\(\therefore \angle RTS = 180^\circ - 90^\circ\)
\(\therefore \angle RTS = 90^\circ\)
In simple words: First, we find \(\angle PRS\) using the exterior angle theorem for \(\triangle PQR\). Since \(\angle TRS\) is the same as \(\angle PRS\), it is \(70^\circ\). Then, using the angle sum property of \(\triangle RTS\), we find \(\angle RTS\).

🎯 Exam Tip: Pay attention to which triangle is being referenced for exterior angles. The exterior angle theorem applies to the *remote* interior angles. When solving for multiple angles, ensure you use the correct angles for each step.

Question 3. In the given figure, bisectors of \(\angle B\) and \(\angle C\) of \(\triangle ABC\) intersect at point P. Prove that \(\angle BPC = 90^\circ + \frac{1}{2} \angle BAC\). Complete the proof by filling in the blanks. (Textbook pg. no.27)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ABC को दर्शाता है, जिसमें कोण B और कोण C के समद्विभाजक बिंदु P पर प्रतिच्छेद करते हैं।
Answer:
Solution:
Proof:
In \(\triangle ABC\),
\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\) ...[Sum of the measures of the angles of a triangle is \(180^\circ\)]
\(\therefore \angle BAC + \angle ABC + \angle ACB = 180^\circ\) ... [Multiplying each term by \(\frac{1}{2}\)]
\(\therefore \frac{1}{2} \angle BAC + \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB = 90^\circ\)
\(\therefore \frac{1}{2} \angle BAC + \angle PBC + \angle PCB = 90^\circ\)
\(\therefore \angle PBC + \angle PCB = 90^\circ - \frac{1}{2} \angle BAC\) .........(i)
In \(\triangle BPC\),
\(\angle BPC + \angle PBC + \angle PCB = 180^\circ\) ....... [Sum of measures of angles of a triangle]
\(\therefore \angle BPC + (90^\circ - \frac{1}{2} \angle BAC) = 180^\circ\) ......[From (i)]
\(\therefore \angle BPC = 180^\circ - 90^\circ + \frac{1}{2} \angle BAC\)
\(\therefore \angle BPC = 90^\circ + \frac{1}{2} \angle BAC\)
In simple words: We start with the angle sum property in \(\triangle ABC\). By multiplying by \(\frac{1}{2}\) and substituting the bisected angles, we find an expression for the sum of \(\angle PBC\) and \(\angle PCB\). Then, we apply the angle sum property in \(\triangle BPC\) and substitute this expression to derive the final formula.

🎯 Exam Tip: This is a standard theorem often proved in class. Practice writing down each step with its corresponding reason (e.g., angle sum property, angle bisector definition). Pay close attention to algebraic manipulations involving fractions.