The bisector of angle A of triangle ABC divides the opposite side at D. Prove that
BD:DC=AB:AC. If AB=4cm and AC=5cm, find triangle ABD:triangle BRC.
For the second part of the sum, the figure would be BA produced to R with RC joined.
I proved the 1st part correctly but couldn\'t prove the second one (traingle ABD:triangle BRC).
Therefore I request you people to please post the second part\'s solution.
Solution
ABC is the triangle AD is the angle bisector of the angle A. D is the point on BC.
To prove BD:DC=AB:AC.
Proof:
Let the perpendicular drawn from A to BC to meet BC at H. Then AH is the height of the
triangle from the base BC. AH is also the height from BD to A. And DC to A, if we consider
ABD and ADC as different triangles .
The area of triangle ABD is give in two ways:
(i)(1/2)the product of two sides and the sine of the angle between the sides, or (ii) (1/2) the
product of base and height.
(1/2)AB*ADsin(A/2) =( 1/2)BC*AH.-----------(1)
Similarly the area of ADC Is given by:
(1/2)AC*AD sin(A/2)=(1/2)DC*AH--------------(2)
Eqality (1)/ Equalty(2) : AB/AC = BC/BD=> BC:BD=AB:AC.
Ratio of Area of triangles: ABD: ACD = (1/2)AB*AD*sin(A/2)/{(1/2)AD*AC*sin(A/2)} =
AB/AC = 4cm/(5cm) = AB:AC =4:5.
Is this helpful?.
ISYU TUNGKOL SA SEKSWLADIDA (ISSUE ABOUT SEXUALITY
The bisector of angle A of triangle ABC divides the opposite side at.pdf
1. The bisector of angle A of triangle ABC divides the opposite side at D. Prove that
BD:DC=AB:AC. If AB=4cm and AC=5cm, find triangle ABD:triangle BRC.
For the second part of the sum, the figure would be BA produced to R with RC joined.
I proved the 1st part correctly but couldn't prove the second one (traingle ABD:triangle BRC).
Therefore I request you people to please post the second part's solution.
Solution
ABC is the triangle AD is the angle bisector of the angle A. D is the point on BC.
To prove BD:DC=AB:AC.
Proof:
Let the perpendicular drawn from A to BC to meet BC at H. Then AH is the height of the
triangle from the base BC. AH is also the height from BD to A. And DC to A, if we consider
ABD and ADC as different triangles .
The area of triangle ABD is give in two ways:
(i)(1/2)the product of two sides and the sine of the angle between the sides, or (ii) (1/2) the
product of base and height.
(1/2)AB*ADsin(A/2) =( 1/2)BC*AH.-----------(1)
Similarly the area of ADC Is given by:
(1/2)AC*AD sin(A/2)=(1/2)DC*AH--------------(2)
Eqality (1)/ Equalty(2) : AB/AC = BC/BD=> BC:BD=AB:AC.
Ratio of Area of triangles: ABD: ACD = (1/2)AB*AD*sin(A/2)/{(1/2)AD*AC*sin(A/2)} =
AB/AC = 4cm/(5cm) = AB:AC =4:5.
Is this helpful?
