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Question 1: In an isosceles triangle $ABC$ with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Show that $OB = OC$.
Solution: Since $AB = AC$, $\angle B = \angle C$. Halves of equal angles are equal, so $\angle OBC = \angle OCB$. In $\triangle OBC$, sides opposite to equal angles are equal, so $OB = OC$.
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Question 2: $AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that $AD$ bisects $BC$.
Solution: In $\triangle ABD$ and $\triangle ACD$: $AB = AC$ (Given), $\angle ADB = \angle ADC = 90^\circ$ (Altitude), $AD = AD$ (Common). By RHS rule, $\triangle ABD \cong \triangle ACD$. Hence $BD = CD$ (CPCT).
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Question 3: $ABC$ is a right angled triangle in which $\angle A = 90^\circ$ and $AB = AC$. Find $\angle B$ and $\angle C$.
Solution: Since $AB = AC$, $\angle B = \angle C$. In $\triangle ABC$, $\angle A + \angle B + \angle C = 180^\circ \Rightarrow 90^\circ + 2\angle B = 180^\circ \Rightarrow 2\angle B = 90^\circ \Rightarrow \angle B = 45^\circ$. So $\angle B = \angle C = 45^\circ$.
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Question 4: Show that the angles of an equilateral triangle are $60^\circ$ each.
Solution: Let $\triangle ABC$ be equilateral. $AB=BC \Rightarrow \angle C = \angle A$. $BC=AC \Rightarrow \angle A = \angle B$. Thus $\angle A = \angle B = \angle C$. Sum is $180^\circ$, so $3\angle A = 180^\circ \Rightarrow \angle A = 60^\circ$.
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Question 5: Triangle $ABC$ is an isosceles triangle with $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$. Show that $\angle BCD$ is a right angle.
Solution: In $\triangle ABC$, $\angle ABC = \angle ACB$. In $\triangle ACD$, $AC=AD$ (since $AB=AC=AD$), so $\angle ACD = \angle ADC$. Sum of angles in $\triangle BCD$: $\angle B + \angle D + (\angle ACB + \angle ACD) = 180^\circ \Rightarrow 2(\angle ACB + \angle ACD) = 180^\circ \Rightarrow \angle BCD = 90^\circ$.
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Question 6: $AB$ 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 $\angle EPA = \angle DPB$. Show that $\triangle DAP \cong \triangle EBP$.
Solution: $\angle EPA + \angle EPD = \angle DPB + \angle EPD \Rightarrow \angle APD = \angle BPE$. In $\triangle DAP$ and $\triangle EBP$: $\angle A = \angle B$ (Given), $AP = BP$ (Midpoint), $\angle APD = \angle BPE$. By ASA rule, $\triangle DAP \cong \triangle EBP$.
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Question 7: 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 $DM = CM$. Show that $\angle DBC$ is a right angle.
Solution: $\triangle AMC \cong \triangle BMD$ (SAS). So $\angle MAC = \angle MBD$ (CPCT), which are alternate angles, so $AC \parallel BD$. Since $\angle ACB = 90^\circ$ and lines are parallel, interior angles sum to $180^\circ$, so $\angle DBC = 90^\circ$.
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Question 8: Show that in a right angled triangle, the hypotenuse is the longest side.
Solution: In right $\triangle ABC$ at $B$, $\angle A + \angle C = 90^\circ$. So $\angle B > \angle A$ and $\angle B > \angle C$. Side opposite to larger angle is longer. $AC$ is opposite to $\angle B$, so $AC > BC$ and $AC > AB$.
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Question 9: In $\triangle ABC$, $\angle A > \angle B$. Show that $BC > AC$.
Solution: The side opposite to the greater angle is longer. Side opposite $\angle A$ is $BC$. Side opposite $\angle B$ is $AC$. Since $\angle A > \angle B$, $BC > AC$.
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Question 10: Prove that the sum of the three altitudes of a triangle is less than the perimeter of the triangle.
Solution: In $\triangle ABD$, $AB > AD$ (Hypotenuse > Altitude). Similarly $BC > BE$ and $AC > CF$. Adding all: $AB + BC + AC > AD + BE + CF$.