Question

in the given fig.if AD⊥BC prove that AB²+CD²=BD²+AC²




THE FIRST ANSWER IS GIVEN BRAINLIEST

Answer 1

Δ ABD
AD² = AB² - BD²

ΔACD = AD² - CD²

AD² = AD²
AB² - BD² = AD² - CD²
AB² + CD² = BD² + AD²

Answer 2

In figure ABC is triangle in which AD⊥BC .

now , we have two right angle triangle .
e.g ∆ADB and ∆ ADC .

for right angle ∆ADB :
we know, according to Pythagoras theorem . if any traingle is a right angle then , it follow
H² = B² + P² { H = hypotenuse, P = perpendicular and B is base of ∆}

so, In ADB ,
AB² = AD² + BD²
AD² = AB² - BD² ------(2)

similarly ∆ADC is a right angle ∆
so, AC² = AD² + CD²
AD² = AC² - CD² -------(2)

from equation (1) and (2)

AB² - BD² = AC² - CD²

AB² + CD² = AC² + BD²
hence proved //