D is the mid-point of side BC of an isosceles triangle ABC . Prove that the circles drawn with either of ...?

Question

the equal sides as a diameter passes through the point D .

PLEASE !!! SOME ONE HELP !!!!

AND NO HATE-ANSWERS PLEASE .

Answer 1

Your answer
ABC is an isosceles triangle
so ab=ac
d is the midpoint of BC
construct the circle and let side ab be its diameter
let another point g other than d pass through the circle(assumed)
construct AG
so angle AGB=90' (diameter property)
so AG is perpendicular to seg BC
therefor AG divides side BC(ABC is isosceles and isosceles triangle property)
so g is midpoint of seg BC
but d is midpoint of seg BC(given)
so d and g are one and the same
so point d passes through the circle drawn keeping ab as the diameter
similarly it can be proved for the other side also

NB i have not written given and to prove please also write them

hope it helps you

Answer 2

utilising comparable triangles ABC and DEC, AB/ DE = BC/ EC (x+7)/ 4 = 3x/ x (x+7)/ 4 = 3 x +7 = 4 multiply 3 = 12 x = 12 - 7 = 5 consequently, x = 5 on an identical time as x =5, AB = x +7 = 5 +7 = 12 BC = 3x = 3(5) = 15 consequently, AB = 12 and BC = 15 in case you're asking what comparable triangles are, they're triangles wherein the corresponding angles are equivalent and their corresponding aspects are in share. i choose this helps :)

Answer 3

Its so simple. triangle ABD and triangle ADC are congruent by using SSS property. This implies angle formed by angle ADB and angle ADC are equal and sum of these is equal to 180degrees. This implies each angle is equal to 90degrees. Hence when a circle is drawn from either of the equal sides as a diameter, point D will pass through it.