# ABC is an obtuse triangle where angle C>90 degrees. AD is perpendicular to BC produced and BE is perpndicular to AC produced prove AB^2=AC*AE+BD*BC

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### 2 Answers

In the question ABC is an obtuse triangle with C> 90 degree. AD is perpendicular to BC produced and BE is perpendicular to AC produced.

Now take the triangle ADC: AD^2 = AC^2 - CD^2

In triangle ABD: AD^2 = AB^2 - DB^2

=> AC^2 - CD^2 = AB^2 - DB^2

=> AB^2 = AC^2 + DB^2 - CD^2 ...(1)

Now take the triangle BCE: BE^2 = BC^2 - CE^2

In triangle BAE: BE^2 = AB^2 - AE^2

=> BC^2 - CE^2 = AB^2 - AE^2

=> AB^2 = BC^2 + AE^2 - CE^2 ...(2)

Adding (1) and (2)

2AB^2 = AC^2 + DB^2 - CD^2 + BC^2 + AE^2 - CE^2

=> 2AB^2 = AC^2 + DB^2 - CD^2 + BC^2 + AE^2 - CE^2

=> 2AB^2 = AC^2 + DB^2 - (DB - CB)^2 + BC^2 + AE^2 - (AE - AC)^2

=> 2AB^2 = AC^2 + DB^2 - (DB^2 +CB^2 - 2*CB*DB)+ BC^2 + AE^2 - (AE^2 +AC^2- 2*AE*AC)

=>2AB^2 = AC^2 + DB^2 - DB^2 -CB^2 + 2*CB*DB+ BC^2 + AE^2 - AE^2 -AC^2+ 2*AE*AC

=> 2AB^2 = AC^2 -AC^2+ DB^2 - DB^2 -CB^2 + BC^2 + 2*CB*DB + AE^2 - AE^2 + 2*AE*AC

=> 2AB^2 = 2*CB*DB + 2*AE*AC

=> AB^2 = BD* BC + AC* AE

**Therefore we have the required result: **

**AB^2 = BD* BC + AC* AE**

First draw the figure.

Consider the right angle ABD. AB is the hypotenuse. So by Pythagoras theorem,

AB^2 = BD^2+AD^2

AB^2= BC(BC+BD)+AD^2

AB^2 = BC*BD +CD(BC+CD)+ AD^2

AB^2 = BC*BD +CD*BC +(CD^2 +AD^2). But in triangle ACD, AC is hypotenuse. So CD^2+AD^2 = AC^2

AB^2 = BC*BD+CD*BC+AC^2.

AB^2= BC*BD + AC*CE +AC^2, As ADEBis concyclic. So BC*CD = AC*CE.

AB^2 = BC*CD+AC(CE+AC). But CE+AC = AC.

AB^2 = BC*CD+AC*AE.