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

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