Question

In the figure, triangle ABC is an obtuse triangle in which angle C > 90°. AD perpendicular to BC produced and BE perpendicular to AC produced. Prove that AB² = BC × BD + AC × AE​

Answer 1

Figure is in the attachment...

Given that triangle ABC is an obtuse triangle in which angle C > 90°. AD perpendicular to BC produced and BE perpendicular to AC produced.

We have to prove ; AB² = BC × BD + AC × AE

Proof

In ∆ACD ; using Pythagoras theorem :

→ AC² = AD² + CD²

→ AD² = AC² - CD²........ (1)

In ∆ABD ; using Pythagoras theorem :

→ AB² = AD² + BD²

→ AD² = AB² - BD² ........(2)

From (1) and (2) :

→ AC² - CD² = AB² - BD²

→ AB² = AC² - CD² + BD² ........ (3)

Now in ∆BCE ; using Pythagoras theorem :

→ BC² = BE² + CE²

→ BE² = BC² - CE² ......... (4)

In ∆BAE ; using Pythagoras theorem :

→ AB² = BE² + AE²

→ BE² = AB² - AE² ........ (5)

Now from (4) and (5) :

→ BC² - CE² = AB² - AE²

→ AB² = BC² - CE² + AE² ....... (6)

Now adding (3) and (6) :

→ AB² + AB² = AC² - CD² + BD² + BC² - CE² + AE²

→ 2AB² = AC² - (BD - BC)² + BD² + BC² - (AE - AC)² + AE²

→ 2AB² = AC² - (BD² - 2.BD.BC + BC²) + BD² + BC² - (AE² - 2.AE.AC + AC²) + AE²

→ 2AB² = AC² - BD² + 2.BD.BC - BC² + BD² + BC² - AE² + 2.AE.AC - AC² + AE²

→ 2AB² = 2.BD.BC + 2.AE.AC

→ 2AB² = 2(BD.BC + AE.AC)

→ AB² = 2(BD × BC + AE × AC)/2

→ AB² = BC × BD + AC × AE

Therefore, L.H.S = R.H.S [Proved]