Question

PA,QB,and RC are each perpendicular to AC where PR and AR intersect at Q.
If AP=x, QB=z,RC=y and BC=b then prove that 1/x+1/y=1/z

Answer 1

In Δ PAC and Δ QBC,

∠ PCA = ∠ QCB   (Common angles)

∠ PAC = ∠ QBC    (Right angle each)

⇒ Δ PAC ~ Δ QBC   (AA similarity)

⇒ PA/QB = AC/BC    (Similar triangles have proportional sides)

⇒ x/y = AC/BC ⇒ y/x = BC/AC .....(1)

In Δ RCA and Δ QBA,

∠ RAC = ∠ QAB    (Common angles)

∠ RCA = ∠ QBA    (Right angle each)

⇒ Δ RCA ~ Δ QBA   (AA similarity)

⇒ RC/QB = AC/AB

⇒ z/y = AC/AB ⇒ y/z = AB/AC .....(2)

Adding (1) and (2), we get

y/x + y/z = (BC+AC)/AC = AC/AC = 1

⇒ y/x + y/z = 1

On multiplying both sides by 1/y, we get

1/x + 1/z = 1/y

Hence proved.