BA is congruent to BC and DA is congruent to EC. Prove BD is congruent to BE.
1 answer:
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Step-by-step explanation:
Notice that, the angle QRS is external to the triangle and adjacent to the angle PRQ. According to the theorem of a external/adjacent angle, we have: m∠QRS = m∠PQR + m∠RPQ, where PQR and RPQ are internal angles.
From the hypothesis, we have:
m∠QRS =(10x−12)∘(10x−12)
m∠PQR = (3x+20)∘(3x+20)
m∠RPQ=(3x−8)∘(3x−8)
Using the first equation and replacing the hypothesis:
m∠QRS = m∠PQR + m∠RPQ
(10x−12)∘(10x−12) = (3x+20)∘(3x+20) + (3x−8)∘(3x−8)
Multiplying and applying the remarkable identity:
Then, we use a calculator to find the roots, which are:
In this case, we will see what root is the right one.
Now, we replace it into m∠QRS =(10x−12)∘(10x−12), because we need to find m∠QRS.
m∠QRS =(10x−12)∘(10x−12) = (10(4.7) - 12) (10(4.7) - 12) = (35) (35) = 1225