Question

Please help........................

Answer 1

y = 2

Explanation:

`\text{If}\ \overrightarrow{BD}\ \text{is the angle bisector of}\ \angle ABC,\ \text{then we have two equations:}\\\\(1)\qquad2x+y=14+y\\(2)\qquad5x-y=x+13y\\------------------\\\\(1)\\2x+y=14+y\qquad\text{subtract}\ y\ \text{from both sides}\\2x=14\qquad\text{divide both sides by 2}\\\boxed{x=7}\\\\\text{Put the value of x to (2):}\\\\5(7)-y=7+13y\\35-y=7+13y\qquad\text{subtract 35 from both sides}\\-y=-28+13y\qquad\text{subtract 13y from both sides}\\-14y=-28\qquad\text{divide both sides by (-14)}\\\boxed{y=2}`

Answer 2

The value of y = 2

Explanation:

∵ BD bisects angle ABC

∴ m∠ABD = m∠CBD⇒ (1)

∵ m∠BAD = 90° , m∠BCD = 90°

∴ m∠BAD = m∠BCD ⇒ (2)

∵ BD is a common side in the two Δ ABD and CBD ⇒ (3)

From (1) , (2) and (3)

∴ The two Δ ABD and CBD are congruent

∴ AD = CD

∵ AD = 5x - y , CD = x + 13y

∴ 5x - y = x + 13y ⇒ collect like terms

∴ 5x - x = 13y + y

∴ 4x = 14y ⇒ (4)

∵ m∠ABD = m∠CBD

∵ m∠ABD = 2x + y , m∠CBD = 14 + y

∴ 2x + y = 14 + y ⇒ 2x + y - y = 14 ⇒ 2x = 14

∴ x = 7 ⇒ substitute the value of x in (4)

∴ 4(7) = 14y

∴ y = 28/14 = 2