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
+ =