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1 of 2

Sol:

From ΔADC, we have

AC²

=

AD²

+

CD²

(Pythagoras Theorem)

(1)

From ΔADB, we have

AB²

=

AD²

+

BD²

(Pythagoras Theorem)

(2)

Subtracting (1) from (2), we have

AB²

–

AC²

=

BD²

–

CD²

AB²

+

CD²

=

BD²

+

AC²

D

A

C

B

Solved Example

If AD ⊥ BC,

prove that AB² + CD² = BD² + AC².

If AD ⊥ BC,

2 of 2

In DPQR, PD ^ QR, such that D lies on QR.

If PQ = a, PR = b, QD = c and DR = d

Prove : (a + b) (a – b) = (c + d) (c – d)

P

b

R

d

D

Q

c

a

Proof :

In DPDQ,

∠PDQ

=

90^o

PD²

+

QD²

=

PQ²

[By Pythagoras theorem]

PD²

+

c²

=

a²

∴

∴

…(i)

In DPDR,

∠PDR

=

90^o

PD²

+

DR²

=

PR²

[By Pythagoras theorem]

PD²

+

d²

=

b²

∴

∴

…(ii)

a²

–

b²

–

d²

∴

[From (i) and (ii)]

c²

=

a²

–

c²

–

d²

∴

b²

=

(a + b)

∴

(a – b)

=

(c + d)

(c – d)

D

PD²

=

a²

–

c²

PD²

=

b²

–

d²