Question

subject mid-point or pythagoras theorem​

Answer 1

Answer:

hey here is your entire proof

pls mark it as brainliest

Step-by-step explanation:

$to \: prove = \\ c {}^{2} = a {}^{2} + b {}^{2} - 2ax \\ \\ so \: considering \: \: △ \: ABD\\ as \: ∠ABD = 90 \\ applying \: pythagoras \: theorem \\ \\ AB {}^{2} = AD {}^{2} + DB {}^{2} \\ \\ c {}^{2} = h {}^{2} + (a - x) {}^{2} \\ \\ h {}^{2} = c {}^{2} - (a - x) {}^{2} \: \: \: \: \: (1)$

$similarly \: \\ considering \: △ \: ADC \\ as \: \: ∠ADC = 90 \\ by \: \: pythagoras \: \: theorem \\ \\ AC {}^{2} = AD {}^{2} + DC {}^{2} \\ \\ b {}^{2} = h {}^{2} + x {}^{2} \\ \\ h {}^{2} = b {}^{2} - x {}^{2} \: \: \: \: \: \: \: \: (2)$

$equating \: (1) \: \: and \: \: (2) \\ \\ c {}^{2} - (a - x) {}^{2} = b {}^{2} - x {}^{2} \\ \\ c {}^{2} - (a {}^{2} + x {}^{2} - 2ax) = b {}^{2} - x {}^{2} \\ \\ c {}^{2} - a {}^{2} - x {}^{2} + 2ax = b {}^{2} - x {}^{2} \\ \\ c {}^{2} = a {}^{2} - 2ax + b {}^{2} \\ \\ c {}^{2} = a {}^{2} + b {}^{2} - 2ax \\ \\ thus \: proved$