ABC and DBC are both isosceles Triangles on a common base BC such that a and d lie on the same side of BC are triangles ADB and ADC congruent which condition do you use if angle BAC is equals to 40 degree and Angle bdc = 100 degree then find angle ADB
Answers
Answer:
- Yes, ∆ ADB ≅ ∆ ADC by Side-side-side congruence criterion.
- ∠ ADB = 130°
Step-by-step explanation:
We are given that, ∆ ABC and ∆ DBC are isosceles triangles and at least two sides of isosceles triangles are equal in length,
∴ BD = DC ….. (i)
And,
AB = AC ….. (ii)
Step 1:
Considering ∆ ADB & ∆ ADC, we have
BD = DC ….. [from (i)]
AB = AC ….. [from (ii)]
AD = AD ….. [common]
∴ By SSS congruence criterion, ∆ ADB ≅ ∆ ADC
∴ ∠BAD = ∠CAD …. [C.P.C.T] ….. (iii)
Also,
∠BAC =∠BAD + ∠CAD
⇒ 40° = 2 * ∠BAD ….. [∵ ∠BAC = 40° (given) and ∠BAD = ∠CAD from (iii)]
⇒ ∠BAD = 40° / 2 = 20° ….. (iv)
Step 2:
Since from (ii), we have AB = AC
∴ ∠ABC = ∠ACB …. [∵ angles opposite to equal sides are equal] …. (v)
In ∆ABC, applying angle sum property,
∠A + ∠ABC +∠ACB = 180°
⇒ 40° + 2* ∠ABC = 180° …. [from (v) and given ∠A = 40° ]
⇒ ∠ABC = 140°/2 = 70°
Also, in ∆ DBC, applying angle sum property,
∠BDC + ∠DBC + ∠DCB = 180°
⇒ 100° + 2* ∠DBC = 180° …. [∵ angles opposite to equal sides BD = CD are equal i.e., ∠DBC = ∠DCB]
⇒ ∠DBC = 80°/2 = 40°
Now,
∠ABD = ∠ABC – ∠DBC
⇒ ∠ABD = 70° – 40° = 30° ….. (vi)
Step 3:
In ∆ ABD, applying angle sum property,
∠BAD + ∠ADB + ∠ABD = 180°
⇒ 20° + ∠ADB + 30° = 180° …. [substituting values from (iv) & (vi)]
⇒ ∠ADB = 180° - 50° = 130°
Answer:
Here you Go ..... hope It Helps
Step-by-step explanation:
Answer:
Yes, ∆ ADB ≅ ∆ ADC by Side-side-side congruence criterion.
∠ ADB = 130°
Step-by-step explanation:
We are given that, ∆ ABC and ∆ DBC are isosceles triangles and at least two sides of isosceles triangles are equal in length,
∴ BD = DC ….. (i)
And,
AB = AC ….. (ii)
Step 1:
Considering ∆ ADB & ∆ ADC, we have
BD = DC ….. [from (i)]
AB = AC ….. [from (ii)]
AD = AD ….. [common]
∴ By SSS congruence criterion, ∆ ADB ≅ ∆ ADC
∴ ∠BAD = ∠CAD …. [C.P.C.T] ….. (iii)
Also,
∠BAC =∠BAD + ∠CAD
⇒ 40° = 2 * ∠BAD ….. [∵ ∠BAC = 40° (given) and ∠BAD = ∠CAD from (iii)]
⇒ ∠BAD = 40° / 2 = 20° ….. (iv)
Step 2:
Since from (ii), we have AB = AC
∴ ∠ABC = ∠ACB …. [∵ angles opposite to equal sides are equal] …. (v)
In ∆ABC, applying angle sum property,
∠A + ∠ABC +∠ACB = 180°
⇒ 40° + 2* ∠ABC = 180° …. [from (v) and given ∠A = 40° ]
⇒ ∠ABC = 140°/2 = 70°
Also, in ∆ DBC, applying angle sum property,
∠BDC + ∠DBC + ∠DCB = 180°
⇒ 100° + 2* ∠DBC = 180° …. [∵ angles opposite to equal sides BD = CD are equal i.e., ∠DBC = ∠DCB]
⇒ ∠DBC = 80°/2 = 40°
Now,
∠ABD = ∠ABC – ∠DBC
⇒ ∠ABD = 70° – 40° = 30° ….. (vi)
Step 3:
In ∆ ABD, applying angle sum property,
∠BAD + ∠ADB + ∠ABD = 180°
⇒ 20° + ∠ADB + 30° = 180° …. [substituting values from (iv) & (vi)]
⇒ ∠ADB = 180° - 50° = 130°