Advertisements
Advertisements
प्रश्न
In the given figure, bisector of ∠BAC intersects side BC at point D. Prove that AB > BD.
उत्तर
Given: Bisector of ∠BAC intersects side BC at point D.
To prove: AB > BD
Proof:
∠BAD ≅ ∠DAC ...(i) ...[Seg AD bisects ∠BAC]
∠ADB is the exterior angle of ∆ADC.
∴ ∠ADB > ∠DAC ...(ii) ...[Property of exterior angle]
∴ ∠ADB > ∠BAD ...(iii) ...[From (i) and (ii)]
In ΔABD,
∠ADB > ∠BAD ...[From (iii)]
∴ AB > BD ...[Side opposite to greater angle is greater]
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, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR.
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)
