ABC and DBC are two Triangles on same base BC such that a and delay on opposite sides of BC if a is equals to AC and BD = to DC show that ad is perpendicular bisector of BC
Answers
Given:
ABC and DBC are two Triangles on the same base BC such that a and delay on opposite sides of BC
AB = AC
BD = DC
To find:
AD is the perpendicular bisector of BC
Construction:
AD intersects BC at O
Solution:
In Δ ABD and Δ ACD, we have
AB = AC ..... [given]
AD = AD ....... [common side]
BD = CD ....... [given]
∴ Δ ABD ≅ Δ ACD ...... [by SSS congruency]
We know that ⇒ CPCT ⇒ this theorem states if two triangles are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other.
We got, Δ ABD ≅ Δ ACD
∴ By CPCT → ∠BAD = ∠CAD ...... (i)
In Δ AOB and Δ AOC, we get
AO = AO ..... [common side]
∠BAD = ∠CAD ...... [from (i)]
AB = AC .... [given]
∴ Δ AOB ≅ Δ AOC ...... [by SSS congruency]
∴ By CPCT → BO = OC and ∠AOB = ∠AOC ..... (ii)
Now, we have
∠AOB + ∠AOC = 180° ..... [Linear pair]
⇒ 2∠AOB = 180° ..... [from (ii) → ∠AOB = ∠AOC]
⇒ ∠AOB =
⇒ ∠AOB = 90°
∴∠AOB = ∠AOC = 90° ...... (iii)
Thus, from (ii) & (iii), we get
BO = OC & ∠AOB = ∠AOC = 90°
⇒ we know that → a perpendicular bisector is a line segment that cuts another line segment into half and at right angles to each other.
⇒ AO ⊥ BC & AO bisects BC
⇒ AD ⊥ BC & AD bisects BC
∴
[Note: Figure is given as an attachment below]
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Answer:
AD is the perpendicular bisector of E
Step-by-step explanation:
ABC and DBC are two Triangles on the same base BC such that a and delay on opposite sides of BC
AB = AC
BD = DC
To find:
AD is the perpendicular bisector of BC
Construction:
AD intersects BC at O
Solution:
In Δ ABD and Δ ACD, we have
AB = AC ..... [given]
AD = AD ....... [common side]
BD = CD ....... [given]
∴ Δ ABD ≅ Δ ACD ...... [by SSS congruency]
We know that ⇒ CPCT ⇒ this theorem states if two triangles are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other.
We got, Δ ABD ≅ Δ ACD
∴ By CPCT → ∠BAD = ∠CAD ...... (i)
In Δ AOB and Δ AOC, we get
AO = AO ..... [common side]
∠BAD = ∠CAD ...... [from (i)]
AB = AC .... [given]
∴ Δ AOB ≅ Δ AOC ...... [by SSS congruency]
∴ By CPCT → BO = OC and ∠AOB = ∠AOC ..... (ii)
Now, we have
∠AOB + ∠AOC = 180° ..... [Linear pair]
⇒ 2∠AOB = 180° ..... [from (ii) → ∠AOB = ∠AOC]
⇒ ∠AOB = \frac{180}{2}
2
180
⇒ ∠AOB = 90°
∴∠AOB = ∠AOC = 90° ...... (iii)
Thus, from (ii) & (iii), we get
BO = OC & ∠AOB = ∠AOC = 90°
⇒ we know that → a perpendicular bisector is a line segment that cuts another line segment into half and at right angles to each other.
⇒ AO ⊥ BC & AO bisects BC
⇒ AD ⊥ BC & AD bisector BC
Therefore AD is the perpendicular bisector of E