In Fig.,PA, QB and RC are each perpendicular to AC. Prove that \(\fra

1.	In Fig.,PA, QB and RC are each perpendicular to AC. Prove that \(\frac{1}{x} +\frac{1}{z} = \frac{1}{y}\).
Answer» We have, PA ⏊ AC, and RC ⏊ AC Let AB = a and BC = b In ΔCQB and ΔCPA ∠QCB = ∠PCA(common) ∠QBC = ∠PCA(Each 90º) Then, ΔCQB ~ ΔCPA (By AA similarity) So,\(\frac{QB}{PA}\) =\(\frac{CB}{CA}\) (Corresponding parts of similar triangle area proportion) Or,\(\frac{y}x\) = \(\frac{b}{{a}+{b}}\) -----------(i) In ΔAQB and ΔARC ∠QAB = ∠RAC(common) ∠ABQ = ∠AQC(Each 90º) Then, ΔAQB ~ ΔARC (By AA similarity) So,\(\frac{QB}{RC}\) = \(\frac{AB}{CA}\) (Corresponding parts of similar triangle area proportion) Or,\(\frac{y}x\) =\(\frac{b}{{a}+{b}}\) -----------(ii) Adding equation i & ii \(\frac{y}{x}+\frac{y}{x} \) = \(\frac{b}{{a}+{b}}\) = \(\frac{a}{{a}+{b}}\) or,\({y}(\frac{1}{x}+\frac{1}{z})\) = \(\frac{b+a}{a+b}\) or,\({y}(\frac{1}{x}+\frac{1}{z})\) = 1 or \(\frac{1}{x}+\frac{1}{z} \) = \(\frac{1}y\)

In Fig.,PA, QB and RC are each perpendicular to AC. Prove that \(\frac{1}{x} +\frac{1}{z} = \frac{1}{y}\).

Answer»

We have,

PA ⏊ AC, and RC ⏊ AC

Let AB = a and BC = b

In ΔCQB and ΔCPA

∠QCB = ∠PCA(common)

∠QBC = ∠PCA(Each 90º)

Then, ΔCQB ~ ΔCPA (By AA similarity)

So,\(\frac{QB}{PA}\) =\(\frac{CB}{CA}\) (Corresponding parts of similar triangle area proportion)

Or,\(\frac{y}x\) = \(\frac{b}{{a}+{b}}\) -----------(i)

In ΔAQB and ΔARC

∠QAB = ∠RAC(common)

∠ABQ = ∠AQC(Each 90º)

Then, ΔAQB ~ ΔARC (By AA similarity)

So,\(\frac{QB}{RC}\) = \(\frac{AB}{CA}\) (Corresponding parts of similar triangle area proportion)

Or,\(\frac{y}x\) =\(\frac{b}{{a}+{b}}\) -----------(ii)

Adding equation i & ii

\(\frac{y}{x}+\frac{y}{x} \) = \(\frac{b}{{a}+{b}}\) = \(\frac{a}{{a}+{b}}\)

or,\({y}(\frac{1}{x}+\frac{1}{z})\) = \(\frac{b+a}{a+b}\)

or,\({y}(\frac{1}{x}+\frac{1}{z})\) = 1

or \(\frac{1}{x}+\frac{1}{z} \) = \(\frac{1}y\)