<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:
19. c 20.b
Step-by-step explanation:
He should use the Pythagorean Theorem to find the missing length.
Since KT is tangent to the circle and TL reaches the center of circle L, the measure of angle LTK is 90 degrees. This means that triangle LTK is a right triangle which means the Pythagorean Theorem can be used.
So,
TL²+(12)²=(13)²
=> TL²+144=169
=> TL²=25
=> TL = 5
Therefore, the radius of circle L is 5 feet.
A² + b² = c²
3² + 4² = c²
9 + 16 = c²
C² = 25
C = 5
so
SM = 5
