In a △ ABC, AD is a median and AE⊥BC. Prove that: AC2=AD2+BC⋅DE+(2BC)2
Get the complete, step-by-step math solution for: "In a △ ABC, AD is a median and AE BC. Prove that: AC^{2}=AD^{2}+BC· DE+(BC)/(2)^{2}". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Apply Pythagorean Theorem in △AEC
Since AE⊥BC, △AEC is a right-angled triangle with the right angle at E. We can apply the Pythagorean theorem, which states that the square of the hypotenuse (AC) is equal to the sum of the squares of the other two sides (AE and EC).
Step 2: Express EC in terms of DC and DE
From the diagram, we can see that the segment EC can be expressed as the difference between DC and DE. Since AD is a median, D is the midpoint of BC, so DC=2BC.
Step 3: Substitute EC and expand the equation
Substitute the expression for EC from the previous step into the Pythagorean equation. Then, expand the term (DC−DE)2 using the algebraic identity (a−b)2=a2−2ab+b2.
Step 4: Apply Pythagorean Theorem in △AED
Similarly, △AED is also a right-angled triangle with the right angle at E. Applying the Pythagorean theorem to this triangle, we get AD2=AE2+DE2.
Step 5: Substitute AD2 and DC=2BC
Now, substitute AD2 for (AE2+DE2) in the expanded equation. Also, replace DC with 2BC since D is the midpoint of BC. This simplifies the equation to the desired form.
Step 6: Rearrange to match the required form
Rearrange the terms to match the exact form required in the problem statement. This completes the proof.