**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

** + = **