Question

in the given figure AD perpendicular to BC prove that AB² + CD² = BD ² + AC ²​

Answer 1

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 //

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Answer 2

Answer:

In the adjoining figure, AB perpendicular BC, AD perpendicular CD and BC = CD. Prove that AC bisects angle BAD.