Question

. ABCD is a rectangle. Points M and N are on BD, such that AM is perpendicular to BD and CN is perpendicular to BD. Prove that
BM^2 + BN^2-DM^2 + DN^2.
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Answer 1

Answer:

6Given:

A rectangle has points M and N on line BD such that AM is perpendicular to BD and CN is perpendicular to BD.

We need to show that BM2 + BN2 = DM2 + DN2.

Based on the provided information, we make the following diagram

Since ΔAMB and ΔCND are right-angled triangles, we use Pythagoras' theorem to see that

AB2 = AM2 + BM2 ........ (1)

CD2 = CN2 + DN2 ........ (2)

Since AB = CD, we can combine Equations (1) and (2) as follows

AM2 + BM2 = CN2 + DN2

AM2 – CN2 = DN2 – BM2 ........ (3)

Following the same reasoning for ΔAMD and ΔCNB, we get the following set of relations

AM2 – CN2 = BN2 – DM2 ........ (4)

Equating Equations (3) and (4), we get

BN2 – DM2 = DN2 – BM2

BM2 + BN2 = DM2 + DN2

Therefore, given the provided information about rectangle ABCD, we see that BM2 + BN2 = DM2 + DN2.

Answer 2

Answer:

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Step-by-step explanation:

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