<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
Answer:
Following are the sample list can be attached as follows:
Step-by-step explanation:
In the given question some information is missing that is attachment of list which can be attached as follows:
please find the attachment list:
by evaluating the list it will give the answer that the both the samples most want to wear mascara the least want to wear lipstick.
Answer:
m = 2
n = 4
Step-by-step explanation:
Ok so to solve this, you want to get it so there is only one variable and then solve that equation. To do this, you can start by doing:
3m + n = 10
multiply both sides by 2
6m + 2n = 20
now, in both of your equations you have 2n so you can add the two equations:
6m + 2n = 20
+ 5m - 2n = 2
11m = 22
divide both sides by 11
m = 2.
Now, plug this value into the original equation, 3m + n = 10:
3 * 2 + n = 10
6 + n = 10
subtract 6 from both sides
n = 4
Answer:
Hope this helps out dude!
Step-by-step explanation:
x = 58°
<u>Explanation:</u>
According to the diagram, lines HI and JK are parallel to each other.
∠AHI = 60°
∠JKI = 62°
∠JKI and ∠HIA are equal because when two lines are parallel then the corresponding angles are equal.
So,
∠JKI = ∠HIA = 62°
In ΔHIA,
∠AHI + ∠HIA + ∠IAH = 180°
60° + 62°+ x = 180°
122° + x = 180°
x = 180° - 122°
x = 58°
