Question

Ab bc pq are perpendicular to bd ab=x cd=y pq=z prove 1/x=1/y=1/z

Answer 1

Refer to the attached image.

Given : $AB\parallel PQ \parallel CD$

To prove: $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$

Proof:

Consider $\Delta DPQ$ and $\Delta DAB$

$\angle D=\angle D$ (Common angle)

$\angle ABD=\angle PQD$ (Each 90 degree)

(By AA criteria, which states "In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.")

Therefore, $\Delta DPQ\sim \Delta DAB$

When triangles are similar then all three pairs of corresponding sides are in the same ratio.

Therefore, $\frac{DP}{DA}=\frac{PQ}{AB}=\frac{DQ}{DB}$

Since, PQ='z' and AB='x'

So, $\frac{DP}{DA}=\frac{DQ}{DB}=\frac{z}{x}$ (Equation 1)

Consider $\Delta BPQ$ and $\Delta BCD$

$\angle B=\angle B$ (Common angle)

$\angle PQB=\angle BDC$ (Each 90 degree)

(By AA criteria, which states "In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.")

Therefore,$\Delta BPQ\sim \Delta BCD$

When triangles are similar then all three pairs of corresponding sides are in the same ratio.

Therefore, $\frac{BP}{BC}=\frac{PQ}{CD}=\frac{BQ}{BD}$

Since, PQ='z' and CD='y'

So, $\frac{BP}{BC}=\frac{BQ}{BD}=\frac{z}{y}$ (Equation 2)

Now,

consider $\frac{z}{x}+\frac{z}{y}$

Substituting values from equation 1 and 2

$\frac{z}{x}+\frac{z}{y}$ = $\frac{DQ}{BD}+\frac{BQ}{BD}$

=$\frac{BD}{BD}$ =1

So, $\frac{z}{x}+\frac{z}{y}=1$

Now, dividing both sides of the equation by 'z'

$\frac{1}{x}+\frac{1}{y}= \frac{1}{z}$

Hence, proved.