Answer:
The proof is explained below.
Step-by-step explanation:
Given m∠ADB = m∠CDB and AD ≅ DC
we have to prove that m∠BAC = m∠BCA and BD⊥ AC
In ΔADO and ΔCDO
∠OAD=∠OCD (∵ADC is an isosceles triangle)
AD=DC (∵Given)
∠ADO=∠CDO (∵Given)
By ASA rule, ΔADO≅ΔCDO
In ΔBAD and ΔBCD
AD=DC (∵ABC is an isosceles triangle)
∠ADB=∠CDB (∵Given)
DB=DB (∵common)
By ASA rule, ΔADB≅ΔCDB
Now, ΔADB≅ΔCDB and ΔADO≅ΔCDO
⇒ ΔADB-ΔADO≅ΔCDB-ΔCDO
⇒ ΔABO≅ΔCBO
Hence, by CPCT, m∠BAC = m∠BCA
Now, we have to prove that BD⊥ AC i.e we have to prove m∠BOA=90°
Now, ΔABO≅ΔCBO therefore by CPCT, m∠BOA = m∠BOC
But, m∠BOA + m∠BOC=180° (linear pair)
⇒ m∠BOA + m∠BOA=180°
⇒ 2m∠BOA=180° ⇒ m∠BOA=90°
Hence, BD⊥ AC
189 (+20), 209 (+40), 249 (+80), 329 (+160), 489 (+320), 809 (+640), 1449...
We are adding twice what we added to the previous number. Each time we double what we're adding.
There are 138 different combinations that they could end up first, second, and third.
We are given the function p(x) = 4x for the perimeter of the square.
Where x represents the known side length of the square.
x is the independent variable and Perimeter P is the dependent variable on x.
<em>Because value of P total depends on the value of side length x of the square.</em>
Therefore, following statement are true about the given function:
<h3>The perimeter is the dependent variable.</h3><h3>The value of p(x) depends on the value of x.</h3><h3>The length of the side of the square is the independent variable.</h3>