Q8 of 45 Page 49

In figure 3.63, seg AD ⊥ seg BC. seg AE is the bisector of ∠CAB and C - E - D. Prove that ∠DAE = 1/2 (∠B - ∠C)

Given: AE is bisector of ∠CAB.


AD is perpendicular to CB


To Prove: ∠DAE = 1/2 (∠B - ∠C)


Proof:


We know that ∠BAE = 1/2 ∠A (1)


∠B + ∠BAD = 90°


∠BAD = 90° - ∠B ……………..(2)


On putting equations (1) and (2)


∠DAE = ∠BAE - ∠BAD


= 1/2 ∠A - (90° - ∠B)


= 1/2 ∠A – 90° + ∠B


= 1/2 ∠A - 1/2 (∠C + ∠A + ∠B) + ∠B


= 1/2 ∠A - 1/2 ∠A - 1/2 ∠B – 1/2 ∠C + ∠B


= 1/2 ∠B – 1/2 ∠C


∴ ∠DAE = 1/2 (∠B - ∠C)


More from this chapter

All 45 →
4

In figure 3.59, point D and E are on side BC of ΔABD, such that BD = CE and AD = AE. Show that ΔABD ≅ ΔACE.

 5

In figure 3.60, point S is any point on side QR of ΔPQR.

Prove that: PQ + QR + RP > 2PS


 6

In figure 3.61, bisector of ∠BAC intersects side BC at point D.

Prove that AB > BD


 7

In figure 3.62, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR