PA, QB and RC are each perpendicular to AC. If AP = x, QB = z, RC = Y, AB = a and BC = b, then prove that  1/x+1/y = 1/z

Q) PA, QB and RC are each perpendicular to AC. If AP = x, QB = z, RC = Y, AB = a and BC = b, then prove that  +   =

Ans: Let’s look at Δ CQB & Δ CPA,

By AA similarity theorem,

∠PAC = ∠QBC  (perpendicular to AC)

∠PCA = ∠QCB (common)

Therefore, Δ CQB ~ Δ CPA

= (sides of similar triangles are proportional to each other)

= ………………(i)

Now in Δ AQB & Δ ARC,

By AA similarity theorem,

∠RCA = ∠QBA  (perpendicular to AC)

∠RAC = ∠QAB (common)

Therefore, Δ AQB ~ Δ ARC

=   (sides of similar triangles are proportional to each other)

= …………….. (ii)

By adding equation (i) and equation (ii), we get

+ = + = = 1

+ =