Q11 of 69 Page 9

In Fig. 9.38, AE bisects ∠CAD and ∠B =∠C. Prove that AE||BC.

Given,

AE bisects ∠CAD


∠B = ∠C


In


∠CAD = ∠B + ∠C


∠CAD = ∠C + ∠C


∠CAD = 2∠C


∠1 + ∠2 = 2∠C (Therefore, ∠CAD = ∠1 + ∠2)


∠2 + ∠2 = 2∠C (Therefore, AE bisects ∠CAD)


2∠2 = 2∠C


∠2 = ∠C (Alternate angles)


Therefore, AE ‖ BC


Hence, proved


More from this chapter

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9

In a Δ ABC, AD bisects ∠A and ∠C >∠B. Prove that ∠ADB >∠ADC.

 10

In Δ ABC, BD ⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° - A.

 12

In Fig. 9.39, AB||DE. Find ∠ACD.

 13

Which of the following statements are true (T) and which are false (F).

(i) Sum of the three angles of a triangle is 180°.


(ii) A triangle can have two right angles.


(iii) All the angles of a triangle can be less than 60°.


(iv) All the angles of a triangle can be greater than 60°.


(v) All the angles of a triangle can be equal to 60°.


(vi) A triangle can have two obtuse angles.


(vii) A triangle can have at most one obtuse angles.


(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.


(ix) An exterior angle of a triangle is led than either of its interior opposite angles.


(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.


(xi) An exterior angle of a triangle is greater than the opposite interior angles.