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NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.1 – Part 4
When two lines are cut by a third line (transversal) co-interior angles are between the pair of lines on the same side of the transversal. If the lines are parallel the co-interior angles are supplementary (add up to 180°) .
Q-7 AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that and (see Fig) . Show that
Solution:
Given, P is mid-point of AB.
- (Adding both sides)
In ,
(P is mid-point of AB)
(Given)
Therefore, By Angle-Side-Angle congruence condition.
- (By corresponding Part of Congruent Triangles) .
Q-8 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. Point D is joined to point B (see Fig) . Show that:
- ΔAMC ≅ ΔBMD
- ∠ DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = 1⟋2 AB
Solution:
Given, right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to point D
M is the mid-point of AB and
- In and
(M is the mid-point)
(Vertically opposite angles)
(Given)
Therefore, (By Side-Angle-Side congruence condition) .
- (By Corresponding Parts of Congruent Triangles) Therefore,
as alternate interior angles are equal.
Now,
(co-interiors angles)
In ΔDBC and ΔACB,
(Common)
(Right angles)
(By Corresponding Parts of Congruent Triangles, already proved)
Therefore (By Side-Angle-Side congruence condition)
(M is mid-point)