From A draw the altitude AH intersecting BC in H
Let's prove that triangle ABH is congruent to triangle ACH
The above 2 triangles are right triangles due to the altitude AH
Angle B= Angle C (given)
Angle AHB = Angle AHC =90° (since AH is the altitude)
Then angle BAH = CAH (both complementary to B & C respectively
And AH is a common side
Now Tri. ABH = Tri. ACH because ASA, hence AB=AC
Answer:
May i ask what the algerbraic question is?
Step-by-step explanation:
Answer:
3(5x+8)
Step-by-step explanation:
Answer:
the answer is 10 or 6^2+8^2=100
√100=10