Internal Angle Bisector Theorem
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In fig, AD is the angle bisector of ∠A. If BD=4cm, DC=3cm and AB=6cm, determine AC.
In △ABC, AB = 3 cm, AC = 4 cm and AD is the bisector of ∠A. Then, BD : DC is —
9:16
16:9
3:4
4:3
- 26 cm
- 35 cm
- 28 cm
- 30 cm
- LN = 9 cm and MN = 15 cm
- LN = 15 cm and MN = 9 cm
- LN = 12 cm and MN = 15 cm
- None of the above
- 3.2 cm
- 2.1 cm
- 2.5 cm
- 5 cm
In △ABC, AB=3 cm, AC=4 cm and AD is the bisector of ∠A. Then, BD=DC.
True
False
D, E, F are the mid points of the sides BC, CA and AB respectively of △ ABC. Then △ DEF is congruent to triangle –
ABC
AEF
BFD, CDE
AFE, BFD, CDE
In △ABC, AB=3 cm, AC=4 cm and AD is the bisector of ∠A. Then, BD:DC is
9:16
4:3
16:9
3:4
D, E, F are the midpoints of the sides BC, CA and AB respectively of an equilateral △ABC. Then △AEF is congruent to triangle –
ABC
AEF
BFD, CDE
AFE, BFD, CDE
The perimeter of △ABC is 49 cm. The angle bisector of ∠A meets BC at M. BM = 6 cm, MC = 8 cm. Find AC.
- 20 cm
- 15 cm
- 12 cm
- 10 cm
- 0.25
- 2.5 cm
- 3.2 cm
- 5 cm
- 2.1 cm
sinxo
- 0.25
- 2.7 cm
- 3.6 cm
- 10.8 cm
- 1.8 cm
(i) Perpendicular bisector of the diameter of a circle passes through the P of the circle.
(ii) If B is image of A in line l and D is image of C in line l, then AC = Q .
(iii) Angle bisector is a ray which divides the angle in R equal parts.
- P-Centre, Q-BD, R-2
- P-Centre, Q-AD, R-1
- P-Centre, Q-AB, R-1
- P-Centre, Q-BC, R-2
If the areas of two similar triangles are equal, prove that they are congruent
- 2.5 cm
- 5 cm
- 2.1 cm
- 3.2 cm
△DGH and △DEF are as shown below.
- 7 units
- 20 units
- 30 units
△DGH and △DEF are as shown below.
- 7 units
- 20 units
- 30 units