Question

A triangular window of a building is shown above. Its diagram represents a ΔABC with ∠A = 90° and AB = AC. Points P and R trisect AB and PQ || RS || AС.

Based on the above, answer the following questions:

1. Show that ΔBPQ ~ ΔBAC.  (1)
2. Prove that PQ = `1/3` AC.  (1)
3. 1. If AB = 3 m, find the lengths of BQ and BS. Verify that BQ = `1/2` BS.  (2)
      OR
   2. Prove that `BR^2 + RS^2 = 4/9 BC^2`.  (2)

Answer 1

i. Given AB = AC, PQ || RS || AC

In ΔBPQ and ΔBAC,

∠B = ∠B  ...(Common angles)

∠BPQ = ∠BAC = 90°   ...(∵ PQ || AC)

Therefore, ΔBPQ ∼ΔBAC by AA similarity.

Hence proved.

ii. Since P and R trisect AB.

∴ BP = `1/3` AB

ΔBPQ ∼ ΔBAC

∴ `(PQ)/(AC) = (BP)/(BA)`

⇒ `(PQ)/(AC) = (1/3 AB)/(AB)`

⇒ `(PQ)/(AC) = 1/3`

Therefore, PQ = `1/3` AC

Hence proved.

iii. a. Given AB = 3 m

AB = AC = 3 m

∠ABC = ∠ACB = 45°  ...(By angle sum property in ΔABC)

In ΔBPQ

cos 45° = `(BP)/(BQ)`

`1/sqrt(2) = (BP)/(BQ)`

BQ = `sqrt(2) BP`

BQ = `sqrt(2)  m   ...(∵ BP = 1/3 AB = 1/3 xx 3 = 1)`

In ΔBRS,

cos 45° = `(BR)/(BS)`

`1/sqrt(2) = (BR)/(BS)`

BS = `sqrt(2) BR   ...(∵ BR = 2/3 AB = 2/3 xx 3 = 2  m)`

BS = `sqrt(2) xx 2`

BS = `2sqrt(2)` m

Verifying

BQ = `sqrt(2)`

2BQ = `2sqrt(2)`

2BQ = BS

BQ = `1/2` BS

Hence verified.

OR

b. Prove that `BR^2 + RS^2 = 4/9 BC^2`

In ΔBRS,

BR = RS

By Pythagoras theorem,

BS² = BR² + SR²

`(sqrt(2)BR)^2 = BR^2 + RS^2`

2BR² = BR² + RS²

⇒ BR² + RS² = 2BR²   ...`("But"  BR = 2/3 AB)`

∴ `BR^2 + RS^2 = 2(2/3 AB)^2`

⇒ `BR^2 + RS^2 = 8/9 AB^2`   ...(1)

In ΔABC

BC² = AB² + AC²

BC² = AB² + AB²   ...(∵ AB = AC)

BC² = 2AВ²

`AB^2 = 1/2 BC^2`

Substitute the value of AB² in equation (1).

`BR^2 + RS^2 = 8/9 (1/2 BC^2)`

`BR^2 + RS^2 = 4/9 BC^2`

Hence proved.