Advertisements
Advertisements
प्रश्न
In the given figure, seg AD ⊥ seg BC. seg AE is the bisector of ∠CAB and C - E - D. Prove that ∠DAE = `1/2` (∠C - ∠B)
उत्तर
Given: seg AD ⊥ seg BC
seg AE is the bisector of ∠CAB
To prove: ∠DAE = `1/2` (∠C – ∠B)
Proof:
∴ ∠CAE = `1/2` ∠A ...(i) [seg AE is the bisector of ∠CAB]
In ∆DAE,
∠DAE + ∠ADE + ∠AED = 180° ...[Sum of the measures of the angles of a triangle is 180°]
∴ ∠DAE + 90° + ∠AED = 180° ...[∵ AD ⊥ BC]
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED ...(ii)
In ∆ACE,
∴ ∠ACE + ∠CAE + ∠AEC = 180° ...[Sum of the measures of the angles of a triangle is 180°]
∠C + ∠A + ∠AED = 180° ...[From (i) and C-D-E]
∴ ∠AED = 180° – ∠C – `1/2` ∠A …(iii)
∴ ∠DAE = 90° – 180° – ∠C + `1/2` ∠A ...[Substituting (iii) in (ii)]
∴ ∠DAE = ∠C + `1/2` ∠A – 90° ...(iv)
In ∆ABC,
∠A + ∠B + ∠C = 180°
∴ `1/2 "∠A" + 1/2 "∠B" + 1/2 "∠C" = 90°` ...[Dividing both side by 2]
∴ `1/2 "∠A" = 90° - 1/2 "∠C" - 1/2 "∠B"` ...(v)
∴ ∠DAE = ∠C + `(90° 1/2 "∠C" - 1/2 "∠B") - 90°` ...[Substituting (v) in (iv)]
∴ ∠DAE = `"∠C" - 1/2 "∠C" - 1/2 "∠B"`
= `1/2 "∠C" - 1/2 "∠B"`
∴ ∠DAE = `1/2`(∠C - ∠B)
APPEARS IN
संबंधित प्रश्न
In the given figure, point A is on the bisector of ∠ XYZ. If AX = 2 cm then find AZ.
In the given figure, ∠RST = 56°, seg PT ⊥ ray ST, seg PR ⊥ ray SR and seg PR ≅ seg PT. Find the measure of ∠RSP. State the reason for your answer.
Prove that, if the bisector of ∠BAC of ΔABC is perpendicular to side BC, then ΔABC is an isosceles triangle.
In ΔPQR, If PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.
In the figure, point D and E are on side BC of ΔABC, such that BD = CE and AD = AE. Show that ΔABD ≅ ΔACE.
In the given figure, bisector of ∠BAC intersects side BC at point D. Prove that AB > BD.
In the given figure, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR.
