Question

In fig if AD perpendicular BC ,prove that AB²+CD²= BD²+AC².
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Answer 1

Answer:

Step-by-step explanation:

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 //

I Hope this answer really helps you...

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Answer 2

SOLUTION:-

Given:

• AD perpendicular BC.

Need to prove:

• AB²+CD²= BD²+AC²

Explanation:

From ∆ ADC,we have

AC² = AD²+CD²---> [By using Pythagoras theorem]

From ∆ ADB,we have

AB² = AD²+BD² --->1 [By using Pythagoras theorem]

On subtracting equation 1 from 2,

We get:

AB² - AC² = BD² - CD²

or, AB² + CD² = BD² + AC²

Hence proved✔️