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Question 1: In a triangle, if the bisector of an angle is perpendicular to the opposite side, prove that the triangle is isosceles.
Solution: Let $AD$ be the bisector of $\angle A$ perpendicular to $BC$. In $\triangle ABD$ and $\triangle ACD$: $\angle BAD = \angle CAD$, $AD = AD$, $\angle ADB = \angle ADC = 90^\circ$. By ASA, $\triangle ABD \cong \triangle ACD$. Thus $AB = AC$.
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Question 2: Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third side.
Solution: Extend median $AD$ to $E$ such that $AD = DE$. Join $EC$. $\triangle ABD \cong \triangle ECD$ (SAS). So $AB = EC$. In $\triangle AEC$, $AC + EC > AE \Rightarrow AC + AB > 2AD$.
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Question 3: $O$ is a point in the interior of a triangle $ABC$. Show that $OA + OB + OC > \frac{1}{2} (AB + BC + CA)$.
Solution: In $\triangle OAB$, $OA + OB > AB$. In $\triangle OBC$, $OB + OC > BC$. In $\triangle OCA$, $OC + OA > CA$. Adding all: $2(OA + OB + OC) > AB + BC + CA$. Divide by 2.
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Question 4: Prove that the perimeter of a triangle is greater than the sum of its three medians.
Solution: We know $AB + AC > 2AD$, $AB + BC > 2BE$, $BC + AC > 2CF$. Adding these: $2(AB + BC + AC) > 2(AD + BE + CF) \Rightarrow AB + BC + AC > AD + BE + CF$.
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Question 5: Two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of triangle $PQR$. Show that $\triangle ABC \cong \triangle PQR$.
Solution: $BC = QR \Rightarrow BM = QN$. In $\triangle ABM$ and $\triangle PQN$: $AB=PQ, AM=PN, BM=QN$. So $\triangle ABM \cong \triangle PQN$ (SSS). Thus $\angle B = \angle Q$. Now in $\triangle ABC$ and $\triangle PQR$: $AB=PQ, \angle B=\angle Q, BC=QR$. So $\triangle ABC \cong \triangle PQR$ (SAS).
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Question 6: If the bisector of the vertical angle of a triangle bisects the base, prove that the triangle is isosceles.
Solution: Let $AD$ be bisector of $\angle A$ and $BD=CD$. Extend $AD$ to $E$ such that $AD=DE$. Join $CE$. $\triangle ABD \cong \triangle ECD$. So $AB=CE$ and $\angle BAD = \angle CED$. Since $\angle BAD = \angle CAD$, $\angle CAD = \angle CED$. So $AC=CE$. Thus $AB=AC$.
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Question 7: $ABC$ is a triangle in which $\angle B = 2\angle C$. $D$ is a point on side $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$. Prove that $\angle BAC = 72^\circ$.
Solution: Construct line $AE$ such that $\angle CAE = \angle C$. Then $AE=EC$. Use exterior angle properties and congruence to show relations leading to $x + 2x + 2x = 180$ or similar logic. (Detailed proof involves construction).
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Question 8: $S$ is any point on side $QR$ of a triangle $PQR$. Show that $PQ + QR + RP > 2PS$.
Solution: In $\triangle PQS$, $PQ + QS > PS$. In $\triangle PRS$, $PR + SR > PS$. Adding both: $PQ + PR + (QS + SR) > 2PS \Rightarrow PQ + PR + QR > 2PS$.
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Question 9: Prove that the difference between any two sides of a triangle is less than the third side.
Solution: In $\triangle ABC$, $AB + BC > AC \Rightarrow AB > AC - BC$. Similarly for other sides. Thus difference is always less than the third side.
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Question 10: In triangle $ABC$, the bisector $AD$ of $\angle A$ is perpendicular to side $BC$. Show that $AB = AC$ and triangle $ABC$ is isosceles.
Solution: In $\triangle ABD$ and $\triangle ACD$: $\angle BAD = \angle CAD$ (Bisector), $AD = AD$ (Common), $\angle ADB = \angle ADC = 90^\circ$ (Perpendicular). By ASA, $\triangle ABD \cong \triangle ACD$. Hence $AB = AC$.