Question

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Answer 1

Given :---

• AB = a
• XY = c
• CD = b
• ∠ ABD = ∠ XYD = ∠ CDB = 90° .

Prove :---

• c(a+b) = ab.

Solution :----

❁❁ Refer To Image First .. ❁❁

it is given that ∠ ABD = ∠ XYD = ∠ CDB = 90°

So, Lines AB, XY and CD are parallel to Each other.

→ AB || XY || CD

Now In ΔABD we have ,

→ AB || XY

So,

→ DY/BD = c/a

→ DY = (c/a)*BD ----------------------- Equation (1)

In Δ BDC we have

→ CD || XY

So,

→ BY/BD = c/b

→ BY = (c/b)*BD ----------------------- Equation (2)

_______________________

Adding Equation(1) and Equation(2) now, we get

→ DY + BY = (c/a)*BD + (c/b)*BD

taking (c * BD) common From LHS, we get,

→ BD = c * BD(1/a + 1/b)

BD will be cancel from both sides ,

→ 1/c = 1/a + 1/b

→ 1/c = (a+b)/ab

→ c(a+b) = ab

✪✪ Hence Proved ✪✪

Answer 2

Answer:

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