In the given fig ab parallel to bc if bd / ad = ad / dc prove that abc is a right angled triangle
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In right ΔADB and ΔADC, we have:
AB² = AD² +BD²
AC² = AD²+ DC²
adding up both
AB² + AC² = 2AD² + BD² + DC²
= 2BD x CD + BD² + CD² as AD x 2 = BD x CD
= (BD + CD)² = BC²
Thus, in ΔABC: AB² + AC² = BC²
Hence, ΔABC is a right triangle and has right angled at A.
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