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/ <---- this will be my angle symbol
/B = /c (equal sides have equal angles opposite to them) as AB= AC
now, ∆Adc and ∆adb
/1 = /2
AC = AB
/C= /B
THERE FORE BY ASA ∆ADC and ∆ADB
ARE CONGRUENT
THEN BY CPCT
CD = BD
AD WILL BE PERPENDICULAR ON BC AS
D IS THE MID POINT OF BC
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/_<---- this will be my angle symbol
/B = /C (equal sides have equal angles opposite to them) as AB= AC
now, ∆ADC and ∆ADB
/_1 = /_2
AC = AB
/_C= /_B
THEREFORE, BY ASA
∆ADC~=∆ADB
THEN BY CPCT
CD = BD
AD WILL BE PERPENDICULAR ON BC AS
D IS THE MID POINT OF BC.
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