# In ΔABC,angle C is obtuse AD perpendicular BCproduced and BE perpendicular AC produced prove tht AB²=BC.BD+AC.AE

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In right angled triangle ABE,

AB² = AE² + BE²

= (AC+CE)² + BE²

= AC² + CE² + 2.AC.CE + BE²

= AC² + (CE²+BE²) + 2.AC.CE (in ΔBCE, CE²+BE² = BC²)

⇒AB² = AC² + BC² + 2.AC.CE --------------------(1)

Similarly in right angled triangle ADB,

AB² = AD² + BD²

= AD² + (BC+CD)²

= AD² + BC² + 2.BC.CD + CD²

= (AD² + CD²) + BC² + 2.BC.CD (In ΔACD, AD²+CD² = AC²)

⇒AB² = AC² + BC² + 2.BC.CD --------------------(2)

From equations (1) and (2);

AC² + BC² + 2.AC.AE = AC² + BC² + 2.BC.CD

⇒ AC.CE = BC.CD ----------------------------------(3)

From triangle ADB,

AB² = AD² + BD²

= (AC² - CD²) + (BC+CD)² (In ΔACD, AC² = AD²+CD²)

= AC² - CD² + BC² + CD² + 2.BC.CD

= AC² + BC² + 2.BC.CD

= AC² + BC² + BC.CD + BC.CD (from equation 3)

= AC² + BC² + BC.CD + AC.CE

= AC² + AC.CE + BC² + BC.CD

= AC(AC+CE) + BC(BC+CD)

⇒**AB² = AC.AE + BC.BD (PROVED)**

In right angled triangle ABE,

AB² = AE² + BE²

= (AC+CE)² + BE²

= AC² + CE² + 2.AC.CE + BE²

= AC² + (CE²+BE²) + 2.AC.CE (in ΔBCE, CE²+BE² = BC²)

⇒AB² = AC² + BC² + 2.AC.CE --------------------(1)

Similarly in right angled triangle ADB,

AB² = AD² + BD²

= AD² + (BC+CD)²

= AD² + BC² + 2.BC.CD + CD²

= (AD² + CD²) + BC² + 2.BC.CD (In ΔACD, AD²+CD² = AC²)

⇒AB² = AC² + BC² + 2.BC.CD --------------------(2)

From equations (1) and (2);

AC² + BC² + 2.AC.AE = AC² + BC² + 2.BC.CD

⇒ AC.CE = BC.CD ----------------------------------(3)

From triangle ADB,

AB² = AD² + BD²

= (AC² - CD²) + (BC+CD)² (In ΔACD, AC² = AD²+CD²)

= AC² - CD² + BC² + CD² + 2.BC.CD

= AC² + BC² + 2.BC.CD

= AC² + BC² + BC.CD + BC.CD (from equation 3)

= AC² + BC² + BC.CD + AC.CE

= AC² + AC.CE + BC² + BC.CD

= AC(AC+CE) + BC(BC+CD)

⇒