Answer:
-"My step-dad's punch packs a wallop!"
-"The stars shining in the night sky were luminous."
-"The two diametrically opposed viewpoints of ketchup and mustard have been on war for decades."
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
Answer: Sometimes
Step-by-step explanation:
This is a very tricky question. We know that in an isosceles triangle, two sides are the same. Because of this, we know that the right angle is always 90 degrees, and the other sides would be 45 degrees each because all triangles add up to 180 degrees. A right triangle could also be scalene, meaning none of the sides are the same, for example a scalene triangle could have sides of 44, 46, and 90 degrees. Rest assured, a right triangle will never be equalateral. Hope this helps :) Merry Christmas
bearing in mind that, whenever we have an absolute value expression, is in effect a piece-wise function with a positive and a negative version of the expression, so
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