Class 9 | Chapter 7 | Triangles | Example 9
Example 9: D is a point on side BC of ∆ ABC such that AD = AC (see Fig. 7.47). Show that AB > AD. Example 8 Class 9 | Home
Read MoreExample 9: D is a point on side BC of ∆ ABC such that AD = AC (see Fig. 7.47). Show that AB > AD. Example 8 Class 9 | Home
Read MoreExample 8: P is a point equidistant from two lines l and m intersecting at point A (see Fig. 7.38). Show that the line AP bisects the angle between them. Example 7 Example 9
Read MoreExample 7: AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see Fig. 7.37). Show that the line PQ is the perpendicular…
Read MoreExample 6: In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (see Fig. 7.29). Show that AD = AE. Example 5 Example 7
Read MoreExample 5: E and F are respectively the mid-points of equal sides AB and AC of ∆ ABC (see Fig. 7.28). Show that BF = CE. Example 4 Example 6
Read MoreExample 4: In ∆ ABC, the bisector AD of ∠ A is perpendicular to side BC (see Fig. 7.27). Show that AB = AC and ∆ ABC is isosceles. Example 3 Example 5
Read MoreExample 3: Line-segment AB is parallel to another line-segment CD. O is the mid-point of AD (see Fig. 7.15). Show that (i) ∆AOB ≅ ∆DOC (ii) O is also the mid-point of BC. Example 2 Example 4
Read MoreExample 2: AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B. Example 1 Example 3
Read MoreExample 1: In Fig. 7.8, OA = OB and OD = OC. Show that (i) ∆ AOD ≅ ∆ BOC and (ii) AD || BC. Class 9 | Home Example 2
Read MoreTheorem 7.8: The sum of any two sides of a triangle is greater than the third side. Theorem 7.7 Class 9 | Home
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