In an obtuse-angled ΔABC (obtuse at B), AD is perpendicular to CB produced. Prove that AC2=AB2+BC2+2BC×BD.
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Step-by-Step Solution
Step 1: Identify Right-Angled Triangles
Given that AD is perpendicular to CB produced, it means that AD forms a right angle with the line segment CB (extended). This creates two right-angled triangles: △ADB and △ADC.
Step 2: Apply Pythagoras Theorem in △ADB
In the right-angled triangle △ADB, AB is the hypotenuse. According to the Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Step 3: Apply Pythagoras Theorem in △ADC
Similarly, in the right-angled triangle △ADC, AC is the hypotenuse. Applying the Pythagoras theorem, we get AC2=AD2+CD2. Let's call this Equation 2.
Step 4: Express CD in terms of BC and BD
From the figure, we can observe that the segment CD is the sum of the segments CB and BD. Since CB is produced to D, C, B, and D are collinear, and B lies between C and D.
Step 5: Substitute CD into Equation 2
Substitute the expression for CD from the previous step into Equation 2. This expands the term CD2 into (CB+BD)2.
Step 6: Expand and Substitute
Expand (CB+BD)2 using the identity (a+b)2=a2+b2+2ab. Then, rearrange the terms and substitute AD2+BD2 with AB2 from Equation 1.
Step 7: Final Proof
By substituting AD2+BD2=AB2 (from Equation 1) into the expanded Equation 2, we arrive at the desired result.