| --- |
| source: ncert |
| topic: geometry |
| class_level: class_9 |
| chapter: ch7 |
| difficulty: intermediate |
| --- |
| |
| Class 9 | Ch7: Triangles (Congruence) |
| CONGRUENCE CRITERIA: |
| SAS: Two sides and included angle equal. |
| ASA: Two angles and included side equal. |
| AAS: Two angles and any corresponding side equal. |
| SSS: All three sides equal. |
| RHS: Right angle, hypotenuse and one side equal (right triangles). |
| IMPORTANT THEOREMS: |
| Theorem 1: Angles opposite equal sides of isosceles triangle are equal. |
| If AB=AC then Angle B=Angle C. |
| Proof: In triangles ABD and ACD (D=midpoint of BC): |
| AB=AC (given), BD=CD (D is midpoint), AD=AD (common). By SSS: triangles congruent. Angle B=Angle C by CPCT. |
| Theorem 2 (Converse): If two angles of triangle are equal, sides opposite them are equal. |
| Theorem 3: In right triangle, hypotenuse is the longest side. |
| Theorem 4: The sum of any two sides of triangle > third side. |
| Theorem 5: Of all line segments drawn from external point to line, perpendicular is shortest. |
| INEQUALITIES IN TRIANGLES: |
| In triangle, greater angle has greater opposite side. |
| If Angle A > Angle B then BC > AC. |
| CPCT: Corresponding Parts of Congruent Triangles. |
| SOLVED EXAMPLES: |
| Example 1: ABC is isosceles with AB=AC. D is midpoint of BC. Prove AD perpendicular to BC. |
| In triangles ABD and ACD: AB=AC, BD=CD, AD=AD. SSS congruence. |
| Angle ADB=Angle ADC (CPCT). Both on straight line, so each=90°. AD perpendicular to BC. |
| Example 2: Prove that triangle ABC where AB=BC=CA (equilateral) has all angles 60°. |
| AB=BC → Angle A=Angle C. BC=CA → Angle A=Angle B. So A=B=C. A+B+C=180°. Each=60°. |
| COMMON MISTAKES: |
| RHS works ONLY for right triangles. Cannot apply to other triangles. |
| CPCT used only AFTER proving congruence. |
| SAS: the angle must be INCLUDED (between the two sides). |
