Question

PLZ HELP WITH THIS QUESTION

Answer 1

Answer:

Given,AD perpendicular to BC

Step-by-step explanation:

Now in ∆ABD

c²=h²+(a-x)²

c²=h²+a²+x²-2ax………(1)

And now ∆ADC

b²=h²+x²………(2)

eq(1)-eq(2) we get

c²-b²=h²+a²+x²-2ax-(h²+x²)

c²-b²=h²+a²+x²-2ax-h²-x²

c²-b²=a²-2ax

c²=a²+b²-2ax

Hence proved.