Given: m∠ ADB = m∠CDB AD ≅ DC Prove: m ∠ BAC = m ∠BCA, MD ⊥ AM

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

Step-by-step explanation:

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

i think it is c, not positive though

step-by-step explanation: