If the angles of 1 triangle are respectively equal to the angles of another triangle prove that the ratio of their corresponding sides is same as the ratio of their corresponding (1) medians (2) angle bisector (3) altitudes
At first this is mid point theorem by altitude
draw two triangle. In first triangle write vertices ABC & draw a altitude from angle A to line BC name it D same thing do with other similar triangle name PQR & draw altitude from P to line QR . I am avoiding that you know altitude form 90 degree solving
given:-
triangle ABC ~ PQR
AB = BC = AC - (1)
RQ QR PR
In triangle ADC & PSR
C = R (since triangle ABC ~ PQR)
D = S = 900 (altitude)
since by AA theorem ,
ADC ~ PSR
AD = AC
PS PR ( B.P.T ) . (2)
By eqn (1) & (2)
AB = BC = AC
PQ QR PR
NOW with median
same diagram but draw median at place of altitude
proof-
BD =DC & QS = SR
AC = BC
PR QR
AC = 2CD => AC = CD cutting 2
PR 2SR PR SR
& C = R
BY SAS Theorem
By angle bisector same as altitude it also done by AA theorem
- thank you
draw two triangle. In first triangle write vertices ABC & draw a altitude from angle A to line BC name it D same thing do with other similar triangle name PQR & draw altitude from P to line QR . I am avoiding that you know altitude form 90 degree solving
given:-
triangle ABC ~ PQR
AB = BC = AC - (1)
RQ QR PR
In triangle ADC & PSR
C = R (since triangle ABC ~ PQR)
D = S = 900 (altitude)
since by AA theorem ,
ADC ~ PSR
AD = AC
PS PR ( B.P.T ) . (2)
By eqn (1) & (2)
AB = BC = AC
PQ QR PR
NOW with median
same diagram but draw median at place of altitude
proof-
BD =DC & QS = SR
AC = BC
PR QR
AC = 2CD => AC = CD cutting 2
PR 2SR PR SR
& C = R
BY SAS Theorem
By angle bisector same as altitude it also done by AA theorem
- thank you