(Q26) AB, CD, PQ are perpendicular to BD. AB = x, CD = y and PQ = z, prove that

1

x

+

1

y

=

1

z

.

Answer :

Given ∠B = ∠Q = ∠D = 90^o

Thus, AB ∥ PQ ∥ CD.

Now in ∆BQP, ∆BDC

∠Q = ∠ (90^o)

∠ = ∠C [∵ Angle Sum property of triangles]

∴ ∆BQP ∼ ∆BDC

(by A.A.A similarity condition)

[∵ Ratio of corresponding sides is equal]

Also in ∆DQP and ∆DBA

∠D = ∠D ( Common )

∠Q = ∠ (90^o)

∴ ∆DQP ∼ ∆DBA (by A.A. similarity condition)

[ Ratio of corresponding sides is equal]

Adding (1) and (2), we get