Question 17: In a parallelogram ABCD, the bisector of ∠A meets DC produced at E. If ∠C = 60°, find ∠AED.

Given: Parallelogram ABCD, ∠C = 60°

To find: ∠AED

Solution:
Since ABCD is a parallelogram, opposite angles are equal.
∠A = ∠C = 60°

The bisector of ∠A meets DC produced at E.
So, ∠AED is the angle formed by the bisector of ∠A and DC.

We know that the sum of the angles of a triangle is 180°.
In triangle AED, we have:
∠AED + ∠DAE + ∠DEA = 180°

Since ∠DAE and ∠DEA are adjacent angles, they add up to ∠A.

Therefore, ∠AED + ∠A = 180°
∠AED + 60° = 180°
∠AED = 180° - 60°
∠AED = 120°

Hence, ∠AED is equal to 120°.

Question 18: In a parallelogram ABCD, if ∠A = 50°, find ∠C and ∠D.

Given: Parallelogram ABCD, ∠A = 50°

To find: ∠C and ∠D

Solution:
In a parallelogram, opposite angles are equal.
So, ∠A = ∠C

Therefore, ∠C = 50°

Also, the sum of the angles in a quadrilateral is 360°.
So, ∠A + ∠B + ∠C + ∠D = 360°

Substituting the values, we have:
50° + ∠B + 50° + ∠D = 360°
∠B + ∠D = 360° - 100°
∠B + ∠D = 260°

Since opposite angles in a parallelogram are equal, we can say that ∠B = ∠D.

So, 2∠B = 260°
∠B = 130°

Therefore, ∠C and ∠D are both equal to 50°.

Question 19: If two parallel lines are intersected by a transversal, the bisectors of two consecutive interior angles are supplementary. Justify this statement.

Solution:
Let's consider two parallel lines, line l1 and line l2, intersected by a transversal line t. Let the two consecutive interior angles be ∠A and ∠B.

From the given information, we know that line l1 || line l2, so ∠A + ∠B = 180° (Alternate Interior Angles Theorem).

Now, let the bisector of ∠A be line m1 and the bisector of ∠B be line m2.

Since m1 is the bisector of ∠A, it divides ∠A into two equal angles, let's call them ∠C and ∠D. Similarly, m2 divides ∠B into two equal angles, let's call them ∠E and ∠F.

Therefore,

I want solutions of exercise 6.3 questions 17 18 19 20