Question

In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then

(a) AQ² + CP² = 2(AC² + PQ²)
(b) 2(AQ² + CP²) = AC² + PQ²
(c) AQ² + CP² = AC² + PQ²
(d) $\mathrm{AQ}+\mathrm{CP}=\frac{1}{2}\left(\mathrm{AC}+\mathrm{PQ}\right)$

Answer 1

Disclaimer: There is mistake in the problem. The question should be "In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and BC respectively, then"

Given: In the right ΔABC, right angled at B. P and Q are points on the sides AB and BC respectively.

Applying Pythagoras theorem,

In ΔAQB,

${\mathrm{AQ}}^2={\mathrm{AB}}^2+{\mathrm{BQ}}^2$ .....(1)

In ΔPBC

.....(2)

Adding (1) and (2), we get

${\mathrm{AQ}}^2+{\mathrm{CP}}^2={\mathrm{AB}}^2+{\mathrm{BQ}}^2+{\mathrm{PB}}^2+{\mathrm{BC}}^2$ .....(3)

In ΔABC,

.....(4)

In ΔPBQ,

${\mathrm{PQ}}^2={\mathrm{PB}}^2+{\mathrm{BQ}}^2$ .....(5)

From (3), (4) and (5), we get

${\mathrm{AQ}}^2+{\mathrm{CP}}^2={\mathrm{AC}}^2+{\mathrm{PQ}}^2$

We got the result as